考虑到移动数字键盘。只能按向上、向左、向右或向下至当前按钮的按钮。您不允许按最下面一行的角按钮（即.*和#）。
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给定一个数N，找出给定长度的可能数。 例如： 对于N=1，可能的数字数量为10（0，1，2，3，…，9） 对于N=2，可能的数字为36 可能的数字：00、0811、12、1422、21、23、25等等。 如果我们从0开始，有效数字将是00，08（计数：2） 如果我们从1开始，有效数字将是11、12、14（计数：3） 如果我们从2开始，有效数字将是22,21,23,25（计数：4） 如果我们从3开始，有效数字将是33、32、36（计数：3） 如果我们从4开始，有效数字将是44,41,45,47（计数：4） 如果我们从5开始，有效数字将是55,54,52,56,58（计数：5） ……………………………… ……………………………… 我们需要打印可能的数字。
N=1是很小的情况，可能的数字是10（0，1，2，3，…，9） 对于N>1，我们需要从某个按钮开始，然后移动到四个方向（上、左、右或下）中的任意一个，该方向指向一个有效的按钮（不应转到*、#）。继续这样做，直到获得N长度编号（深度优先遍历）。
递归解决方案： 移动键盘是4X3的矩形网格（4行3列） 假设Count（i，j，N）表示从位置（i，j）开始的N个长度数字的计数
If N = 1 Count(i, j, N) = 10 Else Count(i, j, N) = Sum of all Count(r, c, N-1) where (r, c) is new position after valid move of length 1 from current position (i, j)
下面是上述递归公式的实现。
C++
// A Naive Recursive C program to count number of possible numbers // of given length #include <stdio.h> // left, up, right, down move from current location int row[] = {0, 0, -1, 0, 1}; int col[] = {0, -1, 0, 1, 0}; // Returns count of numbers of length n starting from key position // (i, j) in a numeric keyboard. int getCountUtil( char keypad[][3], int i, int j, int n) { if (keypad == NULL || n <= 0) return 0; // From a given key, only one number is possible of length 1 if (n == 1) return 1; int k=0, move=0, ro=0, co=0, totalCount = 0; // move left, up, right, down from current location and if // new location is valid, then get number count of length // (n-1) from that new position and add in count obtained so far for (move=0; move<5; move++) { ro = i + row[move]; co = j + col[move]; if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 && keypad[ro][co] != '*' && keypad[ro][co] != '#' ) { totalCount += getCountUtil(keypad, ro, co, n-1); } } return totalCount; } // Return count of all possible numbers of length n // in a given numeric keyboard int getCount( char keypad[][3], int n) { // Base cases if (keypad == NULL || n <= 0) return 0; if (n == 1) return 10; int i=0, j=0, totalCount = 0; for (i=0; i<4; i++) // Loop on keypad row { for (j=0; j<3; j++) // Loop on keypad column { // Process for 0 to 9 digits if (keypad[i][j] != '*' && keypad[i][j] != '#' ) { // Get count when number is starting from key // position (i, j) and add in count obtained so far totalCount += getCountUtil(keypad, i, j, n); } } } return totalCount; } // Driver program to test above function int main( int argc, char *argv[]) { char keypad[4][3] = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }}; printf ( "Count for numbers of length %d: %dn" , 1, getCount(keypad, 1)); printf ( "Count for numbers of length %d: %dn" , 2, getCount(keypad, 2)); printf ( "Count for numbers of length %d: %dn" , 3, getCount(keypad, 3)); printf ( "Count for numbers of length %d: %dn" , 4, getCount(keypad, 4)); printf ( "Count for numbers of length %d: %dn" , 5, getCount(keypad, 5)); return 0; } |
JAVA
// A Naive Recursive Java program // to count number of possible // numbers of given length class GfG { // left, up, right, down // move from current location static int row[] = { 0 , 0 , - 1 , 0 , 1 }; static int col[] = { 0 , - 1 , 0 , 1 , 0 }; // Returns count of numbers of length // n starting from key position // (i, j) in a numeric keyboard. static int getCountUtil( char keypad[][], int i, int j, int n) { if (keypad == null || n <= 0 ) return 0 ; // From a given key, only one // number is possible of length 1 if (n == 1 ) return 1 ; int k = 0 , move = 0 , ro = 0 , co = 0 , totalCount = 0 ; // move left, up, right, down // from current location and if // new location is valid, then // get number count of length // (n-1) from that new position // and add in count obtained so far for (move= 0 ; move< 5 ; move++) { ro = i + row[move]; co = j + col[move]; if (ro >= 0 && ro <= 3 && co >= 0 && co <= 2 && keypad[ro][co] != '*' && keypad[ro][co] != '#' ) { totalCount += getCountUtil(keypad, ro, co, n - 1 ); } } return totalCount; } // Return count of all possible numbers of length n // in a given numeric keyboard static int getCount( char keypad[][], int n) { // Base cases if (keypad == null || n <= 0 ) return 0 ; if (n == 1 ) return 10 ; int i = 0 , j = 0 , totalCount = 0 ; for (i = 0 ; i < 4 ; i++) // Loop on keypad row { for (j= 0 ; j< 3 ; j++) // Loop on keypad column { // Process for 0 to 9 digits if (keypad[i][j] != '*' && keypad[i][j] != '#' ) { // Get count when number is starting from key // position (i, j) and add in count obtained so far totalCount += getCountUtil(keypad, i, j, n); } } } return totalCount; } // Driver code public static void main(String[] args) { char keypad[][] = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }}; System.out.printf( "Count for numbers of" + " length %d: %d" , 1 , getCount(keypad, 1 )); System.out.printf( "Count for numbers of" + "length %d: %d" , 2 , getCount(keypad, 2 )); System.out.printf( "Count for numbers of" + "length %d: %d" , 3 , getCount(keypad, 3 )); System.out.printf( "Count for numbers of" + "length %d: %d" , 4 , getCount(keypad, 4 )); System.out.printf( "Count for numbers of" + "length %d: %d" , 5 , getCount(keypad, 5 )); } } // This code has been contributed by 29AjayKumar |
Python3
# A Naive Recursive Python program to count number of possible numbers # of given length # left, up, right, down move from current location row = [ 0 , 0 , - 1 , 0 , 1 ] col = [ 0 , - 1 , 0 , 1 , 0 ] # Returns count of numbers of length n starting from key position # (i, j) in a numeric keyboard. def getCountUtil(keypad, i, j, n): if (keypad = = None or n < = 0 ): return 0 # From a given key, only one number is possible of length 1 if (n = = 1 ): return 1 k = 0 move = 0 ro = 0 co = 0 totalCount = 0 # move left, up, right, down from current location and if # new location is valid, then get number count of length # (n-1) from that new position and add in count obtained so far for move in range ( 5 ): ro = i + row[move] co = j + col[move] if (ro > = 0 and ro < = 3 and co > = 0 and co < = 2 and keypad[ro][co] ! = '*' and keypad[ro][co] ! = '#' ): totalCount + = getCountUtil(keypad, ro, co, n - 1 ) return totalCount # Return count of all possible numbers of length n # in a given numeric keyboard def getCount(keypad, n): # Base cases if (keypad = = None or n < = 0 ): return 0 if (n = = 1 ): return 10 i = 0 j = 0 totalCount = 0 for i in range ( 4 ): # Loop on keypad row for j in range ( 3 ): # Loop on keypad column # Process for 0 to 9 digits if (keypad[i][j] ! = '*' and keypad[i][j] ! = '#' ): # Get count when number is starting from key # position (i, j) and add in count obtained so far totalCount + = getCountUtil(keypad, i, j, n) return totalCount # Driver code keypad = [[ '1' , '2' , '3' ], [ '4' , '5' , '6' ], [ '7' , '8' , '9' ], [ '*' , '0' , '#' ]] print ( "Count for numbers of length 1:" , getCount(keypad, 1 )) print ( "Count for numbers of length 2:" , getCount(keypad, 2 )) print ( "Count for numbers of length 3:" , getCount(keypad, 3 )) print ( "Count for numbers of length 4:" , getCount(keypad, 4 )) print ( "Count for numbers of length 5:" , getCount(keypad, 5 )) # This code is contributed by subhammahato348 |
C#
// A Naive Recursive C# program // to count number of possible // numbers of given length using System; class GfG { // left, up, right, down // move from current location static int []row = {0, 0, -1, 0, 1}; static int []col = {0, -1, 0, 1, 0}; // Returns count of numbers of length // n starting from key position // (i, j) in a numeric keyboard. static int getCountUtil( char [,]keypad, int i, int j, int n) { if (keypad == null || n <= 0) return 0; // From a given key, only one // number is possible of length 1 if (n == 1) return 1; int k = 0, move = 0, ro = 0, co = 0, totalCount = 0; // move left, up, right, down // from current location and if // new location is valid, then // get number count of length // (n-1) from that new position // and add in count obtained so far for (move=0; move<5; move++) { ro = i + row[move]; co = j + col[move]; if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 && keypad[ro,co] != '*' && keypad[ro,co] != '#' ) { totalCount += getCountUtil(keypad, ro, co, n - 1); } } return totalCount; } // Return count of all possible numbers of length n // in a given numeric keyboard static int getCount( char [,]keypad, int n) { // Base cases if (keypad == null || n <= 0) return 0; if (n == 1) return 10; int i = 0, j = 0, totalCount = 0; for (i = 0; i < 4; i++) // Loop on keypad row { for (j = 0; j < 3; j++) // Loop on keypad column { // Process for 0 to 9 digits if (keypad[i, j] != '*' && keypad[i, j] != '#' ) { // Get count when number is starting from key // position (i, j) and add in count obtained so far totalCount += getCountUtil(keypad, i, j, n); } } } return totalCount; } // Driver code public static void Main() { char [,]keypad = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }}; Console.Write( "Count for numbers of" + " length {0}: {1}" , 1, getCount(keypad, 1)); Console.Write( "Count for numbers of" + "length {0}: {1}" , 2, getCount(keypad, 2)); Console.Write( "Count for numbers of" + "length {0}: {1}" , 3, getCount(keypad, 3)); Console.Write( "Count for numbers of" + "length {0}: {1}" , 4, getCount(keypad, 4)); Console.Write( "Count for numbers of" + "length {0}: {1}" , 5, getCount(keypad, 5)); } } /* This code contributed by PrinciRaj1992 */ |
PHP
<?php // A Naive Recursive PHP program // to count number of possible // numbers of given length // left, up, right, down // move from current location // static $row = array(0, 0, -1, 0, 1); //static $col = array(0, -1, 0, 1, 0); // Returns count of numbers of length // n starting from key position // (i, j) in a numeric keyboard. function getCountUtil( $keypad , $i , $j , $n ) { static $row = array (0,0,-1,0,1); static $col = array (0,-1,0,1,0); if ( $keypad == null || $n <= 0) return 0; // From a given key, only one // number is possible of length 1 if ( $n == 1) return 1; $k = 0; $move = 0; $ro = 0; $co = 0; $totalCount = 0; // move left, up, right, down // from current location and if // new location is valid, then // get number count of length // (n-1) from that new position // and add in count obtained so far for ( $move = 0; $move < 5; $move ++) { $ro = $i + $row [ $move ]; $co = $j + $col [ $move ]; if ( $ro >= 0 && $ro <= 3 && $co >=0 && $co <= 2 && $keypad [ $ro ][ $co ] != '*' && $keypad [ $ro ][ $co ] != '#' ) { $totalCount += getCountUtil( $keypad , $ro , $co , $n - 1); } } return $totalCount ; } // Return count of all possible numbers of length n // in a given numeric keyboard function getCount( $keypad , $n ) { // Base cases if ( $keypad == null || $n <= 0) return 0; if ( $n == 1) return 10; $i = 0; $j = 0; $totalCount = 0; for ( $i = 0; $i < 4; $i ++) // Loop on keypad row { for ( $j = 0; $j < 3; $j ++) // Loop on keypad column { // Process for 0 to 9 digits if ( $keypad [ $i ][ $j ] != '*' && $keypad [ $i ][ $j ] != '#' ) { // Get count when number is starting from key // position (i, j) and add in count obtained so far $totalCount += getCountUtil( $keypad , $i , $j , $n ); } } } return $totalCount ; } // Driver code { $keypad = array ( array ( '1' , '2' , '3' ), array ( '4' , '5' , '6' ), array ( '7' , '8' , '9' ), array ( '*' , '0' , '#' )); echo ( "Count for numbers of" . " length" . getCount( $keypad , 1)); echo ( "Count for numbers of" . " length " . getCount( $keypad , 2)); echo ( "Count for numbers of" . " length " .getCount( $keypad , 3)); echo ( "Count for numbers of" . " length " .getCount( $keypad , 4)); echo ( "Count for numbers of" . " length " .getCount( $keypad , 5)); } // This code has been contributed by Code_Mech. |
Javascript
<script> // A Naive Recursive Javascript program // to count number of possible // numbers of given length // left, up, right, down // move from current location let row=[0, 0, -1, 0, 1]; let col=[0, -1, 0, 1, 0]; // Returns count of numbers of length // n starting from key position // (i, j) in a numeric keyboard. function getCountUtil(keypad,i,j,n) { if (keypad == null || n <= 0) { return 0;} // From a given key, only one // number is possible of length 1 if (n == 1) return 1; let k = 0, move = 0, ro = 0, co = 0, totalCount = 0; // move left, up, right, down // from current location and if // new location is valid, then // get number count of length // (n-1) from that new position // and add in count obtained so far for (move=0; move<5; move++) { ro = i + row[move]; co = j + col[move]; if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 && keypad[ro][co] != '*' && keypad[ro][co] != '#' ) { totalCount += getCountUtil(keypad, ro, co, n - 1); } } return totalCount; } // Return count of all possible numbers of length n // in a given numeric keyboard function getCount(keypad,n) { // Base cases if (keypad == null || n <= 0) return 0; if (n == 1) return 10; let i = 0, j = 0, totalCount = 0; for (i = 0; i < 4; i++) // Loop on keypad row { for (j=0; j<3; j++) // Loop on keypad column { // Process for 0 to 9 digits if (keypad[i][j] != '*' && keypad[i][j] != '#' ) { // Get count when number is starting from key // position (i, j) and add in count obtained so far totalCount += getCountUtil(keypad, i, j, n); } } } return totalCount; } // Driver code let keypad=[[ '1' , '2' , '3' ],[ '4' , '5' , '6' ],[ '7' , '8' , '9' ],[ '*' , '0' , '#' ]]; document.write( "Count for numbers of" + " length " , 1, ": " , getCount(keypad, 1)); document.write( "<br>Count for numbers of" + "length " , 2, ": " , getCount(keypad, 2)); document.write( "<br>Count for numbers of" + "length " , 3, ": " , getCount(keypad, 3)); document.write( "<br>Count for numbers of" + "length " , 4, ": " , getCount(keypad, 4)); document.write( "<br>Count for numbers of" + "length " , 5, ": " , getCount(keypad, 5)); // This code is contributed by avanitrachhadiya2155 </script> |
输出：
Count for numbers of length 1: 10Count for numbers of length 2: 36Count for numbers of length 3: 138Count for numbers of length 4: 532Count for numbers of length 5: 2062
动态规划 在较小的路径上有许多重复的遍历（遍历较小的N）以找到所有可能的较长路径（遍历较大的N）。例如，请参见以下两个图表。在这个遍历中，对于两个起始位置（按钮“4”和“8”）的N=4，我们可以看到N=2的重复遍历很少（例如4->1、6->3、8->9、8->7等）。
由于问题具有两个属性： 最优子结构 和 重叠子问题 利用动态规划可以有效地求解。
下面是实现动态规划的程序。
C++
// A Dynamic Programming based C program to count number of // possible numbers of given length #include <stdio.h> // Return count of all possible numbers of length n // in a given numeric keyboard int getCount( char keypad[][3], int n) { if (keypad == NULL || n <= 0) return 0; if (n == 1) return 10; // left, up, right, down move from current location int row[] = {0, 0, -1, 0, 1}; int col[] = {0, -1, 0, 1, 0}; // taking n+1 for simplicity - count[i][j] will store // number count starting with digit i and length j int count[10][n+1]; int i=0, j=0, k=0, move=0, ro=0, co=0, num = 0; int nextNum=0, totalCount = 0; // count numbers starting with digit i and of lengths 0 and 1 for (i=0; i<=9; i++) { count[i][0] = 0; count[i][1] = 1; } // Bottom up - Get number count of length 2, 3, 4, ... , n for (k=2; k<=n; k++) { for (i=0; i<4; i++) // Loop on keypad row { for (j=0; j<3; j++) // Loop on keypad column { // Process for 0 to 9 digits if (keypad[i][j] != '*' && keypad[i][j] != '#' ) { // Here we are counting the numbers starting with // digit keypad[i][j] and of length k keypad[i][j] // will become 1st digit, and we need to look for // (k-1) more digits num = keypad[i][j] - '0' ; count[num][k] = 0; // move left, up, right, down from current location // and if new location is valid, then get number // count of length (k-1) from that new digit and // add in count we found so far for (move=0; move<5; move++) { ro = i + row[move]; co = j + col[move]; if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 && keypad[ro][co] != '*' && keypad[ro][co] != '#' ) { nextNum = keypad[ro][co] - '0' ; count[num][k] += count[nextNum][k-1]; } } } } } } // Get count of all possible numbers of length "n" starting // with digit 0, 1, 2, ..., 9 totalCount = 0; for (i=0; i<=9; i++) totalCount += count[i][n]; return totalCount; } // Driver program to test above function int main( int argc, char *argv[]) { char keypad[4][3] = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }}; printf ( "Count for numbers of length %d: %dn" , 1, getCount(keypad, 1)); printf ( "Count for numbers of length %d: %dn" , 2, getCount(keypad, 2)); printf ( "Count for numbers of length %d: %dn" , 3, getCount(keypad, 3)); printf ( "Count for numbers of length %d: %dn" , 4, getCount(keypad, 4)); printf ( "Count for numbers of length %d: %dn" , 5, getCount(keypad, 5)); return 0; } |
JAVA
// A Dynamic Programming based Java program to // count number of possible numbers of given length class GFG { // Return count of all possible numbers of length n // in a given numeric keyboard static int getCount( char keypad[][], int n) { if (keypad == null || n <= 0 ) return 0 ; if (n == 1 ) return 10 ; // left, up, right, down move from current location int row[] = { 0 , 0 , - 1 , 0 , 1 }; int col[] = { 0 , - 1 , 0 , 1 , 0 }; // taking n+1 for simplicity - count[i][j] will store // number count starting with digit i and length j int [][]count = new int [ 10 ][n + 1 ]; int i = 0 , j = 0 , k = 0 , move = 0 , ro = 0 , co = 0 , num = 0 ; int nextNum = 0 , totalCount = 0 ; // count numbers starting with digit i // and of lengths 0 and 1 for (i = 0 ; i <= 9 ; i++) { count[i][ 0 ] = 0 ; count[i][ 1 ] = 1 ; } // Bottom up - Get number count of length 2, 3, 4, ... , n for (k = 2 ; k <= n; k++) { for (i = 0 ; i < 4 ; i++) // Loop on keypad row { for (j = 0 ; j < 3 ; j++) // Loop on keypad column { // Process for 0 to 9 digits if (keypad[i][j] != '*' && keypad[i][j] != '#' ) { // Here we are counting the numbers starting with // digit keypad[i][j] and of length k keypad[i][j] // will become 1st digit, and we need to look for // (k-1) more digits num = keypad[i][j] - '0' ; count[num][k] = 0 ; // move left, up, right, down from current location // and if new location is valid, then get number // count of length (k-1) from that new digit and // add in count we found so far for (move = 0 ; move < 5 ; move++) { ro = i + row[move]; co = j + col[move]; if (ro >= 0 && ro <= 3 && co >= 0 && co <= 2 && keypad[ro][co] != '*' && keypad[ro][co] != '#' ) { nextNum = keypad[ro][co] - '0' ; count[num][k] += count[nextNum][k - 1 ]; } } } } } } // Get count of all possible numbers of length "n" // starting with digit 0, 1, 2, ..., 9 totalCount = 0 ; for (i = 0 ; i <= 9 ; i++) totalCount += count[i][n]; return totalCount; } // Driver Code public static void main(String[] args) { char keypad[][] = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }}; System.out.printf( "Count for numbers of length %d: %d" , 1 , getCount(keypad, 1 )); System.out.printf( "Count for numbers of length %d: %d" , 2 , getCount(keypad, 2 )); System.out.printf( "Count for numbers of length %d: %d" , 3 , getCount(keypad, 3 )); System.out.printf( "Count for numbers of length %d: %d" , 4 , getCount(keypad, 4 )); System.out.printf( "Count for numbers of length %d: %d" , 5 , getCount(keypad, 5 )); } } // This code is contributed by Rajput-Ji |
Python3
# A Dynamic Programming based C program to count number of # possible numbers of given length # Return count of all possible numbers of length n # in a given numeric keyboard def getCount(keypad, n): if (keypad = = None or n < = 0 ): return 0 if (n = = 1 ): return 10 # left, up, right, down move from current location row = [ 0 , 0 , - 1 , 0 , 1 ] col = [ 0 , - 1 , 0 , 1 , 0 ] # taking n+1 for simplicity - count[i][j] will store # number count starting with digit i and length j # count[10][n+1] count = [[ 0 ] * (n + 1 )] * 10 i = 0 j = 0 k = 0 move = 0 ro = 0 co = 0 num = 0 nextNum = 0 totalCount = 0 # count numbers starting with # digit i and of lengths 0 and 1 for i in range ( 10 ): count[i][ 0 ] = 0 count[i][ 1 ] = 1 # Bottom up - Get number # count of length 2, 3, 4, ... , n for k in range ( 2 , n + 1 ): for i in range ( 4 ): # Loop on keypad row for j in range ( 3 ): # Loop on keypad column # Process for 0 to 9 digits if (keypad[i][j] ! = '*' and keypad[i][j] ! = '#' ): # Here we are counting the numbers starting with # digit keypad[i][j] and of length k keypad[i][j] # will become 1st digit, and we need to look for # (k-1) more digits num = ord (keypad[i][j]) - 48 count[num][k] = 0 # move left, up, right, down from current location # and if new location is valid, then get number # count of length (k-1) from that new digit and # add in count we found so far for move in range ( 5 ): ro = i + row[move] co = j + col[move] if (ro > = 0 and ro < = 3 and co > = 0 and co < = 2 and keypad[ro][co] ! = '*' and keypad[ro][co] ! = '#' ): nextNum = ord (keypad[ro][co]) - 48 count[num][k] + = count[nextNum][k - 1 ] # Get count of all possible numbers of length "n" starting # with digit 0, 1, 2, ..., 9 totalCount = 0 for i in range ( 10 ): totalCount + = count[i][n] return totalCount # Driver code if __name__ = = "__main__" : keypad = [[ '1' , '2' , '3' ], [ '4' , '5' , '6' ], [ '7' , '8' , '9' ], [ '*' , '0' , '#' ]] print ( "Count for numbers of length" , 1 , ":" , getCount(keypad, 1 )) print ( "Count for numbers of length" , 2 , ":" , getCount(keypad, 2 )) print ( "Count for numbers of length" , 3 , ":" , getCount(keypad, 3 )) print ( "Count for numbers of length" , 4 , ":" , getCount(keypad, 4 )) print ( "Count for numbers of length" , 5 , ":" , getCount(keypad, 5 )) # This code is contributed by subhammahato348 |
C#
// A Dynamic Programming based C# program to // count number of possible numbers of given Length using System; class GFG { // Return count of all possible numbers of Length n // in a given numeric keyboard static int getCount( char [,]keypad, int n) { if (keypad == null || n <= 0) return 0; if (n == 1) return 10; // left, up, right, down move // from current location int []row = {0, 0, -1, 0, 1}; int []col = {0, -1, 0, 1, 0}; // taking n+1 for simplicity - count[i,j] // will store number count starting with // digit i and.Length j int [,]count = new int [10,n + 1]; int i = 0, j = 0, k = 0, move = 0, ro = 0, co = 0, num = 0; int nextNum = 0, totalCount = 0; // count numbers starting with digit i // and of.Lengths 0 and 1 for (i = 0; i <= 9; i++) { count[i, 0] = 0; count[i, 1] = 1; } // Bottom up - Get number count of // Length 2, 3, 4, ... , n for (k = 2; k <= n; k++) { for (i = 0; i < 4; i++) // Loop on keypad row { for (j = 0; j < 3; j++) // Loop on keypad column { // Process for 0 to 9 digits if (keypad[i, j] != '*' && keypad[i, j] != '#' ) { // Here we are counting the numbers starting with // digit keypad[i,j] and of.Length k keypad[i,j] // will become 1st digit, and we need to look for // (k-1) more digits num = keypad[i, j] - '0' ; count[num, k] = 0; // move left, up, right, down from current location // and if new location is valid, then get number // count of.Length (k-1) from that new digit and //.Add in count we found so far for (move = 0; move < 5; move++) { ro = i + row[move]; co = j + col[move]; if (ro >= 0 && ro <= 3 && co >= 0 && co <= 2 && keypad[ro, co] != '*' && keypad[ro, co] != '#' ) { nextNum = keypad[ro, co] - '0' ; count[num, k] += count[nextNum, k - 1]; } } } } } } // Get count of all possible numbers of.Length "n" // starting with digit 0, 1, 2, ..., 9 totalCount = 0; for (i = 0; i <= 9; i++) totalCount += count[i, n]; return totalCount; } // Driver Code public static void Main(String[] args) { char [,]keypad = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }}; Console.Write( "Count for numbers of.Length {0}: {1}" , 1, getCount(keypad, 1)); Console.Write( "Count for numbers of.Length {0}: {1}" , 2, getCount(keypad, 2)); Console.Write( "Count for numbers of.Length {0}: {1}" , 3, getCount(keypad, 3)); Console.Write( "Count for numbers of.Length {0}: {1}" , 4, getCount(keypad, 4)); Console.Write( "Count for numbers of.Length {0}: {1}" , 5, getCount(keypad, 5)); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // A Dynamic Programming based Javascript program to // count number of possible numbers of given length // Return count of all possible numbers of length n // in a given numeric keyboard function getCount(keypad, n) { if (keypad == null || n <= 0) return 0; if (n == 1) return 10; // left, up, right, down move // from current location let row = [ 0, 0, -1, 0, 1 ]; let col = [ 0, -1, 0, 1, 0 ]; // Taking n+1 for simplicity - count[i][j] // will store number count starting with // digit i and length j let count = new Array(10); for (let i = 0; i < 10; i++) { count[i] = new Array(n + 1); for (let j = 0; j < n + 1; j++) { count[i][j] = 0; } } let i = 0, j = 0, k = 0, move = 0, ro = 0, co = 0, num = 0; let nextNum = 0, totalCount = 0; // count numbers starting with digit i // and of lengths 0 and 1 for (i = 0; i <= 9; i++) { count[i][0] = 0; count[i][1] = 1; } // Bottom up - Get number count // of length 2, 3, 4, ... , n for (k = 2; k <= n; k++) { // Loop on keypad row for (i = 0; i < 4; i++) { // Loop on keypad column for (j = 0; j < 3; j++) { // Process for 0 to 9 digits if (keypad[i][j] != '*' && keypad[i][j] != '#' ) { // Here we are counting the numbers starting with // digit keypad[i][j] and of length k keypad[i][j] // will become 1st digit, and we need to look for // (k-1) more digits num = keypad[i][j].charCodeAt(0) - '0' .charCodeAt(0); count[num][k] = 0; // Move left, up, right, down from current location // and if new location is valid, then get number // count of length (k-1) from that new digit and // add in count we found so far for (move = 0; move < 5; move++) { ro = i + row[move]; co = j + col[move]; if (ro >= 0 && ro <= 3 && co >= 0 && co <= 2 && keypad[ro][co] != '*' && keypad[ro][co] != '#' ) { nextNum = keypad[ro][co].charCodeAt(0) - '0' .charCodeAt(0); count[num][k] += count[nextNum][k - 1]; } } } } } } // Get count of all possible numbers of length "n" // starting with digit 0, 1, 2, ..., 9 totalCount = 0; for (i = 0; i <= 9; i++) totalCount += count[i][n]; return totalCount; } // Driver Code let keypad = [ [ '1' , '2' , '3' ], [ '4' , '5' , '6' ], [ '7' , '8' , '9' ], [ '*' , '0' , '#' ] ]; document.write( "Count for numbers of length " + 1 + " : " + getCount(keypad, 1) + "<br>" ) document.write( "Count for numbers of length " + 2 + " : " + getCount(keypad, 2) + "<br>" ) document.write( "Count for numbers of length " + 3 + " : " + getCount(keypad, 3) + "<br>" ) document.write( "Count for numbers of length " + 4 + " : " + getCount(keypad, 4) + "<br>" ) document.write( "Count for numbers of length " + 5 + " : " + getCount(keypad, 5) + "<br>" ) // This code is contributed by rag2127 </script> |
输出：
Count for numbers of length 1: 10Count for numbers of length 2: 36Count for numbers of length 3: 138Count for numbers of length 4: 532Count for numbers of length 5: 2062
空间优化解决方案： 上述动态规划方法也在O（n）时间内运行，并且需要O（n）辅助空间，因为只有一个用于循环运行n次，另一个用于循环运行恒定时间。我们可以看到，第n次迭代只需要第（n-1）次迭代中的数据，所以我们不需要保留旧迭代中的数据。我们可以用两个大小为10的数组来实现一种节省空间的动态规划方法。感谢Nik提出这个解决方案。
C++
// A Space Optimized C++ program to count number of possible numbers // of given length #include <bits/stdc++.h> using namespace std; // Return count of all possible numbers of length n // in a given numeric keyboard int getCount( char keypad[][3], int n) { if (keypad == NULL || n <= 0) return 0; if (n == 1) return 10; // odd[i], even[i] arrays represent count of numbers starting // with digit i for any length j int odd[10], even[10]; int i = 0, j = 0, useOdd = 0, totalCount = 0; for (i = 0; i <= 9; i++) odd[i] = 1; // for j = 1 for (j = 2; j <= n; j++) // Bottom Up calculation from j = 2 to n { useOdd = 1 - useOdd; // Here we are explicitly writing lines for each number 0 // to 9. But it can always be written as DFS on 4X3 grid // using row, column array valid moves if (useOdd == 1) { even[0] = odd[0] + odd[8]; even[1] = odd[1] + odd[2] + odd[4]; even[2] = odd[2] + odd[1] + odd[3] + odd[5]; even[3] = odd[3] + odd[2] + odd[6]; even[4] = odd[4] + odd[1] + odd[5] + odd[7]; even[5] = odd[5] + odd[2] + odd[4] + odd[8] + odd[6]; even[6] = odd[6] + odd[3] + odd[5] + odd[9]; even[7] = odd[7] + odd[4] + odd[8]; even[8] = odd[8] + odd[0] + odd[5] + odd[7] + odd[9]; even[9] = odd[9] + odd[6] + odd[8]; } else { odd[0] = even[0] + even[8]; odd[1] = even[1] + even[2] + even[4]; odd[2] = even[2] + even[1] + even[3] + even[5]; odd[3] = even[3] + even[2] + even[6]; odd[4] = even[4] + even[1] + even[5] + even[7]; odd[5] = even[5] + even[2] + even[4] + even[8] + even[6]; odd[6] = even[6] + even[3] + even[5] + even[9]; odd[7] = even[7] + even[4] + even[8]; odd[8] = even[8] + even[0] + even[5] + even[7] + even[9]; odd[9] = even[9] + even[6] + even[8]; } } // Get count of all possible numbers of length "n" starting // with digit 0, 1, 2, ..., 9 totalCount = 0; if (useOdd == 1) { for (i = 0; i <= 9; i++) totalCount += even[i]; } else { for (i = 0; i <= 9; i++) totalCount += odd[i]; } return totalCount; } // Driver program to test above function int main() { char keypad[4][3] = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }}; cout << "Count for numbers of length 1: " << getCount(keypad, 1) << endl; cout << "Count for numbers of length 2: " << getCount(keypad, 2) << endl; cout << "Count for numbers of length 3: " << getCount(keypad, 3) << endl; cout << "Count for numbers of length 4: " << getCount(keypad, 4) << endl; cout << "Count for numbers of length 5: " << getCount(keypad, 5) << endl; return 0; } //This code is contributed by Mayank Tyagi |
C
// A Space Optimized C program to count number of possible numbers // of given length #include <stdio.h> // Return count of all possible numbers of length n // in a given numeric keyboard int getCount( char keypad[][3], int n) { if (keypad == NULL || n <= 0) return 0; if (n == 1) return 10; // odd[i], even[i] arrays represent count of numbers starting // with digit i for any length j int odd[10], even[10]; int i = 0, j = 0, useOdd = 0, totalCount = 0; for (i=0; i<=9; i++) odd[i] = 1; // for j = 1 for (j=2; j<=n; j++) // Bottom Up calculation from j = 2 to n { useOdd = 1 - useOdd; // Here we are explicitly writing lines for each number 0 // to 9. But it can always be written as DFS on 4X3 grid // using row, column array valid moves if (useOdd == 1) { even[0] = odd[0] + odd[8]; even[1] = odd[1] + odd[2] + odd[4]; even[2] = odd[2] + odd[1] + odd[3] + odd[5]; even[3] = odd[3] + odd[2] + odd[6]; even[4] = odd[4] + odd[1] + odd[5] + odd[7]; even[5] = odd[5] + odd[2] + odd[4] + odd[8] + odd[6]; even[6] = odd[6] + odd[3] + odd[5] + odd[9]; even[7] = odd[7] + odd[4] + odd[8]; even[8] = odd[8] + odd[0] + odd[5] + odd[7] + odd[9]; even[9] = odd[9] + odd[6] + odd[8]; } else { odd[0] = even[0] + even[8]; odd[1] = even[1] + even[2] + even[4]; odd[2] = even[2] + even[1] + even[3] + even[5]; odd[3] = even[3] + even[2] + even[6]; odd[4] = even[4] + even[1] + even[5] + even[7]; odd[5] = even[5] + even[2] + even[4] + even[8] + even[6]; odd[6] = even[6] + even[3] + even[5] + even[9]; odd[7] = even[7] + even[4] + even[8]; odd[8] = even[8] + even[0] + even[5] + even[7] + even[9]; odd[9] = even[9] + even[6] + even[8]; } } // Get count of all possible numbers of length "n" starting // with digit 0, 1, 2, ..., 9 totalCount = 0; if (useOdd == 1) { for (i=0; i<=9; i++) totalCount += even[i]; } else { for (i=0; i<=9; i++) totalCount += odd[i]; } return totalCount; } // Driver program to test above function int main() { char keypad[4][3] = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' } }; printf ( "Count for numbers of length %d: %dn" , 1, getCount(keypad, 1)); printf ( "Count for numbers of length %d: %dn" , 2, getCount(keypad, 2)); printf ( "Count for numbers of length %d: %dn" , 3, getCount(keypad, 3)); printf ( "Count for numbers of length %d: %dn" , 4, getCount(keypad, 4)); printf ( "Count for numbers of length %d: %dn" , 5, getCount(keypad, 5)); return 0; } |
JAVA
// A Space Optimized Java program to // count number of possible numbers // of given length class GFG { // Return count of all possible numbers of // length n in a given numeric keyboard static int getCount( char keypad[][], int n) { if (keypad == null || n <= 0 ) return 0 ; if (n == 1 ) return 10 ; // odd[i], even[i] arrays represent count of // numbers starting with digit i for any length j int []odd = new int [ 10 ]; int []even = new int [ 10 ]; int i = 0 , j = 0 , useOdd = 0 , totalCount = 0 ; for (i = 0 ; i <= 9 ; i++) odd[i] = 1 ; // for j = 1 // Bottom Up calculation from j = 2 to n for (j = 2 ; j <= n; j++) { useOdd = 1 - useOdd; // Here we are explicitly writing lines // for each number 0 to 9. But it can always be // written as DFS on 4X3 grid using row, // column array valid moves if (useOdd == 1 ) { even[ 0 ] = odd[ 0 ] + odd[ 8 ]; even[ 1 ] = odd[ 1 ] + odd[ 2 ] + odd[ 4 ]; even[ 2 ] = odd[ 2 ] + odd[ 1 ] + odd[ 3 ] + odd[ 5 ]; even[ 3 ] = odd[ 3 ] + odd[ 2 ] + odd[ 6 ]; even[ 4 ] = odd[ 4 ] + odd[ 1 ] + odd[ 5 ] + odd[ 7 ]; even[ 5 ] = odd[ 5 ] + odd[ 2 ] + odd[ 4 ] + odd[ 8 ] + odd[ 6 ]; even[ 6 ] = odd[ 6 ] + odd[ 3 ] + odd[ 5 ] + odd[ 9 ]; even[ 7 ] = odd[ 7 ] + odd[ 4 ] + odd[ 8 ]; even[ 8 ] = odd[ 8 ] + odd[ 0 ] + odd[ 5 ] + odd[ 7 ] + odd[ 9 ]; even[ 9 ] = odd[ 9 ] + odd[ 6 ] + odd[ 8 ]; } else { odd[ 0 ] = even[ 0 ] + even[ 8 ]; odd[ 1 ] = even[ 1 ] + even[ 2 ] + even[ 4 ]; odd[ 2 ] = even[ 2 ] + even[ 1 ] + even[ 3 ] + even[ 5 ]; odd[ 3 ] = even[ 3 ] + even[ 2 ] + even[ 6 ]; odd[ 4 ] = even[ 4 ] + even[ 1 ] + even[ 5 ] + even[ 7 ]; odd[ 5 ] = even[ 5 ] + even[ 2 ] + even[ 4 ] + even[ 8 ] + even[ 6 ]; odd[ 6 ] = even[ 6 ] + even[ 3 ] + even[ 5 ] + even[ 9 ]; odd[ 7 ] = even[ 7 ] + even[ 4 ] + even[ 8 ]; odd[ 8 ] = even[ 8 ] + even[ 0 ] + even[ 5 ] + even[ 7 ] + even[ 9 ]; odd[ 9 ] = even[ 9 ] + even[ 6 ] + even[ 8 ]; } } // Get count of all possible numbers of // length "n" starting with digit 0, 1, 2, ..., 9 totalCount = 0 ; if (useOdd == 1 ) { for (i = 0 ; i <= 9 ; i++) totalCount += even[i]; } else { for (i = 0 ; i <= 9 ; i++) totalCount += odd[i]; } return totalCount; } // Driver Code public static void main(String[] args) { char keypad[][] = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }}; System.out.printf( "Count for numbers of length %d: %d" , 1 , getCount(keypad, 1 )); System.out.printf( "Count for numbers of length %d: %d" , 2 , getCount(keypad, 2 )); System.out.printf( "Count for numbers of length %d: %d" , 3 , getCount(keypad, 3 )); System.out.printf( "Count for numbers of length %d: %d" , 4 , getCount(keypad, 4 )); System.out.printf( "Count for numbers of length %d: %d" , 5 , getCount(keypad, 5 )); } } // This code is contributed by PrinciRaj1992 |
Python3
# A Space Optimized Python program to count # number of possible numbers # of given length # Return count of all possible numbers # of length n # in a given numeric keyboard def getCount(keypad, n): if ( not keypad or n < = 0 ): return 0 if (n = = 1 ): return 10 # odd[i], even[i] arrays represent # count of numbers starting # with digit i for any length j odd = [ 0 ] * 10 even = [ 0 ] * 10 i = 0 j = 0 useOdd = 0 totalCount = 0 for i in range ( 10 ): odd[i] = 1 # for j = 1 for j in range ( 2 ,n + 1 ): # Bottom Up calculation from j = 2 to n useOdd = 1 - useOdd # Here we are explicitly writing lines for each number 0 # to 9. But it can always be written as DFS on 4X3 grid # using row, column array valid moves if (useOdd = = 1 ): even[ 0 ] = odd[ 0 ] + odd[ 8 ] even[ 1 ] = odd[ 1 ] + odd[ 2 ] + odd[ 4 ] even[ 2 ] = odd[ 2 ] + odd[ 1 ] + odd[ 3 ] + odd[ 5 ] even[ 3 ] = odd[ 3 ] + odd[ 2 ] + odd[ 6 ] even[ 4 ] = odd[ 4 ] + odd[ 1 ] + odd[ 5 ] + odd[ 7 ] even[ 5 ] = odd[ 5 ] + odd[ 2 ] + odd[ 4 ] + odd[ 8 ] + odd[ 6 ] even[ 6 ] = odd[ 6 ] + odd[ 3 ] + odd[ 5 ] + odd[ 9 ] even[ 7 ] = odd[ 7 ] + odd[ 4 ] + odd[ 8 ] even[ 8 ] = odd[ 8 ] + odd[ 0 ] + odd[ 5 ] + odd[ 7 ] + odd[ 9 ] even[ 9 ] = odd[ 9 ] + odd[ 6 ] + odd[ 8 ] else : odd[ 0 ] = even[ 0 ] + even[ 8 ] odd[ 1 ] = even[ 1 ] + even[ 2 ] + even[ 4 ] odd[ 2 ] = even[ 2 ] + even[ 1 ] + even[ 3 ] + even[ 5 ] odd[ 3 ] = even[ 3 ] + even[ 2 ] + even[ 6 ] odd[ 4 ] = even[ 4 ] + even[ 1 ] + even[ 5 ] + even[ 7 ] odd[ 5 ] = even[ 5 ] + even[ 2 ] + even[ 4 ] + even[ 8 ] + even[ 6 ] odd[ 6 ] = even[ 6 ] + even[ 3 ] + even[ 5 ] + even[ 9 ] odd[ 7 ] = even[ 7 ] + even[ 4 ] + even[ 8 ] odd[ 8 ] = even[ 8 ] + even[ 0 ] + even[ 5 ] + even[ 7 ] + even[ 9 ] odd[ 9 ] = even[ 9 ] + even[ 6 ] + even[ 8 ] # Get count of all possible numbers of length "n" starting # with digit 0, 1, 2, ..., 9 totalCount = 0 if (useOdd = = 1 ): for i in range ( 10 ): totalCount + = even[i] else : for i in range ( 10 ): totalCount + = odd[i] return totalCount # Driver program to test above function if __name__ = = "__main__" : keypad = [[ '1' , '2' , '3' ], [ '4' , '5' , '6' ], [ '7' , '8' , '9' ], [ '*' , '0' , '#' ]] print ( "Count for numbers of length " , 1 , ": " , getCount(keypad, 1 )) print ( "Count for numbers of length " , 2 , ": " , getCount(keypad, 2 )) print ( "Count for numbers of length " , 3 , ": " , getCount(keypad, 3 )) print ( "Count for numbers of length " , 4 , ": " , getCount(keypad, 4 )) print ( "Count for numbers of length " , 5 , ": " , getCount(keypad, 5 )) # This code is contributed by # ChitraNayal |
C#
// A Space Optimized C# program to // count number of possible numbers // of given length using System; class GFG { // Return count of all possible numbers of // length n in a given numeric keyboard static int getCount( char [,]keypad, int n) { if (keypad == null || n <= 0) return 0; if (n == 1) return 10; // odd[i], even[i] arrays represent count of // numbers starting with digit i for any length j int []odd = new int [10]; int []even = new int [10]; int i = 0, j = 0, useOdd = 0, totalCount = 0; for (i = 0; i <= 9; i++) odd[i] = 1; // for j = 1 // Bottom Up calculation from j = 2 to n for (j = 2; j <= n; j++) { useOdd = 1 - useOdd; // Here we are explicitly writing lines // for each number 0 to 9. But it can always be // written as DFS on 4X3 grid using row, // column array valid moves if (useOdd == 1) { even[0] = odd[0] + odd[8]; even[1] = odd[1] + odd[2] + odd[4]; even[2] = odd[2] + odd[1] + odd[3] + odd[5]; even[3] = odd[3] + odd[2] + odd[6]; even[4] = odd[4] + odd[1] + odd[5] + odd[7]; even[5] = odd[5] + odd[2] + odd[4] + odd[8] + odd[6]; even[6] = odd[6] + odd[3] + odd[5] + odd[9]; even[7] = odd[7] + odd[4] + odd[8]; even[8] = odd[8] + odd[0] + odd[5] + odd[7] + odd[9]; even[9] = odd[9] + odd[6] + odd[8]; } else { odd[0] = even[0] + even[8]; odd[1] = even[1] + even[2] + even[4]; odd[2] = even[2] + even[1] + even[3] + even[5]; odd[3] = even[3] + even[2] + even[6]; odd[4] = even[4] + even[1] + even[5] + even[7]; odd[5] = even[5] + even[2] + even[4] + even[8] + even[6]; odd[6] = even[6] + even[3] + even[5] + even[9]; odd[7] = even[7] + even[4] + even[8]; odd[8] = even[8] + even[0] + even[5] + even[7] + even[9]; odd[9] = even[9] + even[6] + even[8]; } } // Get count of all possible numbers of // length "n" starting with digit 0, 1, 2, ..., 9 totalCount = 0; if (useOdd == 1) { for (i = 0; i <= 9; i++) totalCount += even[i]; } else { for (i = 0; i <= 9; i++) totalCount += odd[i]; } return totalCount; } // Driver Code public static void Main(String[] args) { char [,]keypad = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }}; Console.Write( "Count for numbers of length {0}: {1}" , 1, getCount(keypad, 1)); Console.Write( "Count for numbers of length {0}: {1}" , 2, getCount(keypad, 2)); Console.Write( "Count for numbers of length {0}: {1}" , 3, getCount(keypad, 3)); Console.Write( "Count for numbers of length {0}: {1}" , 4, getCount(keypad, 4)); Console.Write( "Count for numbers of length {0}: {1}" , 5, getCount(keypad, 5)); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // A Space Optimized javascript program to // count number of possible numbers // of given length // Return count of all possible numbers of // length n in a given numeric keyboard function getCount(keypad , n) { if (keypad == null || n <= 0) return 0; if (n == 1) return 10; // odd[i], even[i] arrays represent count of // numbers starting with digit i for any length j var odd = Array.from({length: 10}, (_, i) => 0); var even = Array.from({length: 10}, (_, i) => 0); var i = 0, j = 0, useOdd = 0, totalCount = 0; for (i = 0; i <= 9; i++) odd[i] = 1; // for j = 1 // Bottom Up calculation from j = 2 to n for (j = 2; j <= n; j++) { useOdd = 1 - useOdd; // Here we are explicitly writing lines // for each number 0 to 9. But it can always be // written as DFS on 4X3 grid using row, // column array valid moves if (useOdd == 1) { even[0] = odd[0] + odd[8]; even[1] = odd[1] + odd[2] + odd[4]; even[2] = odd[2] + odd[1] + odd[3] + odd[5]; even[3] = odd[3] + odd[2] + odd[6]; even[4] = odd[4] + odd[1] + odd[5] + odd[7]; even[5] = odd[5] + odd[2] + odd[4] + odd[8] + odd[6]; even[6] = odd[6] + odd[3] + odd[5] + odd[9]; even[7] = odd[7] + odd[4] + odd[8]; even[8] = odd[8] + odd[0] + odd[5] + odd[7] + odd[9]; even[9] = odd[9] + odd[6] + odd[8]; } else { odd[0] = even[0] + even[8]; odd[1] = even[1] + even[2] + even[4]; odd[2] = even[2] + even[1] + even[3] + even[5]; odd[3] = even[3] + even[2] + even[6]; odd[4] = even[4] + even[1] + even[5] + even[7]; odd[5] = even[5] + even[2] + even[4] + even[8] + even[6]; odd[6] = even[6] + even[3] + even[5] + even[9]; odd[7] = even[7] + even[4] + even[8]; odd[8] = even[8] + even[0] + even[5] + even[7] + even[9]; odd[9] = even[9] + even[6] + even[8]; } } // Get count of all possible numbers of // length "n" starting with digit 0, 1, 2, ..., 9 totalCount = 0; if (useOdd == 1) { for (i = 0; i <= 9; i++) totalCount += even[i]; } else { for (i = 0; i <= 9; i++) totalCount += odd[i]; } return totalCount; } // Driver Code var keypad = [[ '1' , '2' , '3' ], [ '4' , '5' , '6' ], [ '7' , '8' , '9' ], [ '*' , '0' , '#' ]]; document.write( "Count for numbers of length " + 1+ ": " + getCount(keypad, 1)); document.write( "<br>Count for numbers of length " + 2+ ": " + getCount(keypad, 2)); document.write( "<br>Count for numbers of length " + 3+ ": " + getCount(keypad, 3)); document.write( "<br>Count for numbers of length " + 4+ ": " + getCount(keypad, 4)); document.write( "<br>Count for numbers of length " + 5+ ": " + getCount(keypad, 5)); // This code is contributed by 29AjayKumar </script> |
输出：
Count for numbers of length 1: 10Count for numbers of length 2: 36Count for numbers of length 3: 138Count for numbers of length 4: 532Count for numbers of length 5: 2062
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