位旋转：旋转（或循环移位）是一种类似于移位的操作，只是一端脱落的位被放回另一端。 在向左旋转时，从左端脱落的钻头被放回右端。 在右旋转中，从右端脱落的钻头放回左端。
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例子： 设n用8位存储。n=11100101向左旋转3会使n=00101111（左移3位，前3位放回最后一位）。如果使用16位或32位存储n，则n（000…11100101）的左旋转变为00。。00 11100101 0 如果使用8位存储n，则n=11100101的右转为3表示n=10111100（右移3位，最后3位放回第一位）。如果n是使用16位或32位存储的，则n（000…11100101）的右旋转为3 101 000..00 11100 .
C++
// C++ code to rotate bits // of number #include<iostream> using namespace std; #define INT_BITS 32 class gfg { /*Function to left rotate n by d bits*/ public : int leftRotate( int n, unsigned int d) { /* In n<<d, last d bits are 0. To put first 3 bits of n at last, do bitwise or of n<<d with n >>(INT_BITS - d) */ return (n << d)|(n >> (INT_BITS - d)); } /*Function to right rotate n by d bits*/ int rightRotate( int n, unsigned int d) { /* In n>>d, first d bits are 0. To put last 3 bits of at first, do bitwise or of n>>d with n <<(INT_BITS - d) */ return (n >> d)|(n << (INT_BITS - d)); } }; /* Driver code*/ int main() { gfg g; int n = 16; int d = 2; cout << "Left Rotation of " << n << " by " << d << " is " ; cout << g.leftRotate(n, d); cout << "Right Rotation of " << n << " by " << d << " is " ; cout << g.rightRotate(n, d); getchar (); } // This code is contributed by SoM15242 |
C
#include<stdio.h> #define INT_BITS 32 /*Function to left rotate n by d bits*/ int leftRotate( int n, unsigned int d) { /* In n<<d, last d bits are 0. To put first 3 bits of n at last, do bitwise or of n<<d with n >>(INT_BITS - d) */ return (n << d)|(n >> (INT_BITS - d)); } /*Function to right rotate n by d bits*/ int rightRotate( int n, unsigned int d) { /* In n>>d, first d bits are 0. To put last 3 bits of at first, do bitwise or of n>>d with n <<(INT_BITS - d) */ return (n >> d)|(n << (INT_BITS - d)); } /* Driver program to test above functions */ int main() { int n = 16; int d = 2; printf ( "Left Rotation of %d by %d is " , n, d); printf ( "%d" , leftRotate(n, d)); printf ( "Right Rotation of %d by %d is " , n, d); printf ( "%d" , rightRotate(n, d)); getchar (); } |
JAVA
// Java code to rotate bits // of number class GFG { static final int INT_BITS = 32 ; /*Function to left rotate n by d bits*/ static int leftRotate( int n, int d) { /* In n<<d, last d bits are 0. To put first 3 bits of n at last, do bitwise or of n<<d with n >>(INT_BITS - d) */ return (n << d) | (n >> (INT_BITS - d)); } /*Function to right rotate n by d bits*/ static int rightRotate( int n, int d) { /* In n>>d, first d bits are 0. To put last 3 bits of at first, do bitwise or of n>>d with n <<(INT_BITS - d) */ return (n >> d) | (n << (INT_BITS - d)); } // Driver code public static void main(String arg[]) { int n = 16 ; int d = 2 ; System.out.print( "Left Rotation of " + n + " by " + d + " is " ); System.out.print(leftRotate(n, d)); System.out.print( "Right Rotation of " + n + " by " + d + " is " ); System.out.print(rightRotate(n, d)); } } // This code is contributed by Anant Agarwal. |
蟒蛇3
# Python3 code to # rotate bits of number INT_BITS = 32 # Function to left # rotate n by d bits def leftRotate(n, d): # In n<<d, last d bits are 0. # To put first 3 bits of n at # last, do bitwise or of n<<d # with n >>(INT_BITS - d) return (n << d)|(n >> (INT_BITS - d)) # Function to right # rotate n by d bits def rightRotate(n, d): # In n>>d, first d bits are 0. # To put last 3 bits of at # first, do bitwise or of n>>d # with n <<(INT_BITS - d) return (n >> d)|(n << (INT_BITS - d)) & 0xFFFFFFFF # Driver program to # test above functions n = 16 d = 2 print ( "Left Rotation of" ,n, "by" ,d, "is" ,end = " " ) print (leftRotate(n, d)) print ( "Right Rotation of" ,n, "by" ,d, "is" ,end = " " ) print (rightRotate(n, d)) # This code is contributed by # Smitha Dinesh Semwal |
C#
// C# program to rotate // bits of a number using System; class GFG { static int INT_BITS = 32; /* Function to left rotate n by d bits*/ static int leftRotate( int n, int d) { /* In n<<d, last d bits are 0. To put first 3 bits of n at last, do bitwise or of n<<d with n >>(INT_BITS - d) */ return (n << d) | (n >> (INT_BITS - d)); } /*Function to right rotate n by d bits*/ static int rightRotate( int n, int d) { /* In n>>d, first d bits are 0. To put last 3 bits of at first, do bitwise or of n>>d with n <<(INT_BITS - d) */ return (n >> d) | (n << (INT_BITS - d)); } // Driver code public static void Main() { int n = 16; int d = 2; Console.Write( "Left Rotation of " + n + " by " + d + " is " ); Console.Write(leftRotate(n, d)); Console.Write( "Right Rotation of " + n + " by " + d + " is " ); Console.Write(rightRotate(n, d)); } } // This code is contributed by Sam007 |
Javascript
<script> // JavaScript code to rotate bits // of number let INT_BITS = 32; /*Function to left rotate n by d bits*/ function leftRotate( n, d) { /* In n<<d, last d bits are 0. To put first 3 bits of n at last, do bitwise or of n<<d with n >>(INT_BITS - d) */ return (n << d)|(n >> (INT_BITS - d)); } /*Function to right rotate n by d bits*/ function rightRotate( n, d) { /* In n>>d, first d bits are 0. To put last 3 bits of at first, do bitwise or of n>>d with n <<(INT_BITS - d) */ return (n >> d)|(n << (INT_BITS - d)); } /* Driver code*/ let n = 16; let d = 2; document.write( "Left Rotation of " + n + " by " + d + " is " ); document.write(leftRotate(n, d)); document.write( "<br>" ); document.write( "Right Rotation of " + n + " by " + d + " is " ); document.write(rightRotate(n, d)); </script> |
输出：
Left Rotation of 16 by 2 is 64Right Rotation of 16 by 2 is 4
时间复杂性： O（1）
辅助空间： O（1）
对于16位数字：
C++
#include <bits/stdc++.h> using namespace std; #define SHORT_SIZE 16 // function to rotate the given unsigned short // in the left direction unsigned short leftRotate(unsigned short x, short d) { /** * By doing x << d, we move the first(right most) d bits * to the left most d bits, and at the same time we move * the left most d bits to the right side, * performing OR operation between the two gives use the * required result. * */ return (x << d) | (x >> (SHORT_SIZE - d)); } // function to rotate the given unsigned short // in the right direction unsigned short rightRotate(unsigned short x, short d) { /** * By doing x >> d, we move the first(left most) d bits * to the right most d bits, and at the same time we move * the right most d bits to the right side, * performing OR operation between the two gives use the * required result. * */ return (x >> d) | (x << (SHORT_SIZE - d)); } /* Driver program to test above functions */ int main() { // Test case unsigned short n = 28; short d = 2; cout << leftRotate(n, d) << endl; cout << rightRotate(n, d) << endl; return 0; } // This code is contributed by ganesh227 |
C
#include <stdio.h> #define SHORT_SIZE 16 // function to rotate the given unsigned short // in the left direction unsigned short leftRotate(unsigned short x, short d) { /** * By doing x << d, we move the first(right most) d bits * to the left most d bits, and at the same time we move * the left most d bits to the right side, * performing OR operation between the two gives use the * required result. * */ return (x << d) | (x >> (SHORT_SIZE - d)); } // function to rotate the given unsigned short // in the right direction unsigned short rightRotate(unsigned short x, short d) { /** * By doing x >> d, we move the first(left most) d bits * to the right most d bits, and at the same time we move * the right most d bits to the right side, * performing OR operation between the two gives use the * required result. * */ return (x >> d) | (x << (SHORT_SIZE - d)); } /* Driver program to test above functions */ int main() { // Test case unsigned short n = 28; short d = 2; printf ( "%d" , leftRotate(n, d)); printf ( "%d" , rightRotate(n, d)); return 0; } // This code is contributed by ganesh227 |
JAVA
/*package whatever //do not write package name here */ import java.io.*; class GFG { public static void main (String[] args) { int N= 28 ; int D= 2 ; rotate(N,D); } static void rotate( int N, int D) { // your code here int t= 16 ; int left= ((N<<D) | N>>(t-D)) & 0xFFFF ; int right=((N>>D) | N<<(t-D)) & 0xFFFF ; System.out.println(left); System.out.println(right); } } |
蟒蛇3
SHORT_SIZE = 16 # function to rotate the given unsigned short # in the left direction def leftRotate(x, d): return (x << d) | (x >> (SHORT_SIZE - d)) # function to rotate the given unsigned short # in the right direction def rightRotate(x, d): return (x >> d) | (x << (SHORT_SIZE - d)) & 0xDDDDDF # Driver program to test above functions # Test case n = 28 d = 2 print ( "Left Rotation of" ,n, "by" ,d, "is" ,end = " " ) print (leftRotate(n, d)) print ( "Right Rotation of" ,n, "by" ,d, "is" ,end = " " ) print (rightRotate(n, d)) # This code is contributed by shivanisinghss2110 |
C#
/*package whatever //do not write package name here */ using System; using System.Collections.Generic; public class GFG { public static void Main(String[] args) { int N = 28; int D = 2; rotate(N, D); } // Driver code static void rotate( int N, int D) { int t = 16; int left = ((N<<D) | N>>(t-D)) & 0xFFFF; int right = ((N>>D) | N<<(t-D)) & 0xFFFF; Console.WriteLine(left); Console.WriteLine(right); } } // This code is contributed by umadevi9616 |
Javascript
<script> function rotate(N,D) { // your code here let t = 16; let left = ((N<<D) | N>>(t-D)) & 0xFFFF; let right = ((N>>D) | N<<(t-D)) & 0xFFFF; document.write(left + "<br>" ); document.write(right + "<br>" ); } let N = 28; let D = 2; rotate(N, D); // This code is contributed by avanitrachhadiya2155 </script> |
Left Rotation of 28 by 2 is 112Right Rotation of 28 by 2 is 7
时间复杂性 ：O（1）
空间复杂性 ：O（1）
如果您在上述程序中发现任何错误或其他解决相同问题的方法，请写评论。
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