给定一个数字x和x的二进制表示中的两个位置（从右侧），编写一个函数，在给定的两个位置交换n位并返回结果。还指出这两组位不重叠。
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方法1 设p1和p2为两个给定位置。
例1
Input:x = 47 (00101111)p1 = 1 (Start from the second bit from the right side)p2 = 5 (Start from the 6th bit from the right side)n = 3 (No of bits to be swapped)Output:227 (11100011)The 3 bits starting from the second bit (from the right side) are swapped with 3 bits starting from 6th position (from the right side)
例2
Input:x = 28 (11100)p1 = 0 (Start from first bit from right side)p2 = 3 (Start from 4th bit from right side)n = 2 (No of bits to be swapped)Output:7 (00111)The 2 bits starting from 0th position (from right side) areswapped with 2 bits starting from 4th position (from right side)
解决方案 我们需要交换两组比特。XOR的使用方式与以前类似 交换2个号码 .以下是算法。
1) Move all bits of the first set to the rightmost side set1 = (x >> p1) & ((1U << n) - 1)Here the expression (1U << n) - 1 gives a number that contains last n bits set and other bits as 0. We do & with this expression so that bits other than the last n bits become 0.2) Move all bits of second set to rightmost side set2 = (x >> p2) & ((1U << n) - 1)3) XOR the two sets of bits xor = (set1 ^ set2) 4) Put the xor bits back to their original positions. xor = (xor << p1) | (xor << p2)5) Finally, XOR the xor with original number so that the two sets are swapped. result = x ^ xor
实施：
C++
// C++ Program to swap bits // in a given number #include <bits/stdc++.h> using namespace std; int swapBits(unsigned int x, unsigned int p1, unsigned int p2, unsigned int n) { /* Move all bits of first set to rightmost side */ unsigned int set1 = (x >> p1) & ((1U << n) - 1); /* Move all bits of second set to rightmost side */ unsigned int set2 = (x >> p2) & ((1U << n) - 1); /* Xor the two sets */ unsigned int Xor = (set1 ^ set2); /* Put the Xor bits back to their original positions */ Xor = (Xor << p1) | (Xor << p2); /* Xor the 'Xor' with the original number so that the two sets are swapped */ unsigned int result = x ^ Xor; return result; } /* Driver code*/ int main() { int res = swapBits(28, 0, 3, 2); cout << "Result = " << res; return 0; } // This code is contributed by rathbhupendra |
C
// C Program to swap bits // in a given number #include <stdio.h> int swapBits(unsigned int x, unsigned int p1, unsigned int p2, unsigned int n) { /* Move all bits of first set to rightmost side */ unsigned int set1 = (x >> p1) & ((1U << n) - 1); /* Move all bits of second set to rightmost side */ unsigned int set2 = (x >> p2) & ((1U << n) - 1); /* XOR the two sets */ unsigned int xor = (set1 ^ set2); /* Put the xor bits back to their original positions */ xor = (xor << p1) | (xor << p2); /* XOR the 'xor' with the original number so that the two sets are swapped */ unsigned int result = x ^ xor; return result; } /* Driver program to test above function*/ int main() { int res = swapBits(28, 0, 3, 2); printf ( "Result = %d " , res); return 0; } |
JAVA
// Java Program to swap bits // in a given number class GFG { static int swapBits( int x, int p1, int p2, int n) { // Move all bits of first set // to rightmost side int set1 = (x >> p1) & (( 1 << n) - 1 ); // Move all bits of second set // to rightmost side int set2 = (x >> p2) & (( 1 << n) - 1 ); // XOR the two sets int xor = (set1 ^ set2); // Put the xor bits back to // their original positions xor = (xor << p1) | (xor << p2); // XOR the 'xor' with the original number // so that the two sets are swapped int result = x ^ xor; return result; } // Driver program public static void main(String[] args) { int res = swapBits( 28 , 0 , 3 , 2 ); System.out.println( "Result = " + res); } } // This code is contributed by prerna saini. |
Python3
# Python program to # swap bits in a given number def swapBits(x, p1, p2, n): # Move all bits of first # set to rightmost side set1 = (x >> p1) & (( 1 << n) - 1 ) # Move all bits of second # set to rightmost side set2 = (x >> p2) & (( 1 << n) - 1 ) # XOR the two sets xor = (set1 ^ set2) # Put the xor bits back # to their original positions xor = (xor << p1) | (xor << p2) # XOR the 'xor' with the # original number so that the # two sets are swapped result = x ^ xor return result # Driver code res = swapBits( 28 , 0 , 3 , 2 ) print ( "Result =" , res) # This code is contributed # by Anant Agarwal. |
C#
// C# Program to swap bits // in a given number using System; class GFG { static int swapBits( int x, int p1, int p2, int n) { // Move all bits of first // set to rightmost side int set1 = (x >> p1) & ((1 << n) - 1); // Move all bits of second set // set to rightmost side int set2 = (x >> p2) & ((1 << n) - 1); // XOR the two sets int xor = (set1 ^ set2); // Put the xor bits back to // their original positions xor = (xor << p1) | (xor << p2); // XOR the 'xor' with the original number // so that the two sets are swapped int result = x ^ xor; return result; } // Driver program public static void Main() { int res = swapBits(28, 0, 3, 2); Console.WriteLine( "Result = " + res); } } // This code is contributed by vt_m. |
PHP
<?php // PHP Program to swap bits // in a given number // function returns // the swapped bits function swapBits( $x , $p1 , $p2 , $n ) { // Move all bits of first // set to rightmost side $set1 = ( $x >> $p1 ) & ((1 << $n ) - 1); // Move all bits of second // set to rightmost side $set2 = ( $x >> $p2 ) & ((1 << $n ) - 1); // XOR the two sets $xor = ( $set1 ^ $set2 ); // Put the xor bits back to // their original positions $xor = ( $xor << $p1 ) | ( $xor << $p2 ); // XOR the 'xor' with the // original number so that // the two sets are swapped $result = $x ^ $xor ; return $result ; } // Driver Code $res = swapBits(28, 0, 3, 2); echo "Result = " , $res ; // This code is contributed by anuj_67. ?> |
Javascript
<script> // Javascript Program to swap bits // in a given number function swapBits(x, p1, p2, n) { // Move all bits of first set // to rightmost side let set1 = (x >> p1) & ((1 << n) - 1); // Move all bits of second set // to rightmost side let set2 = (x >> p2) & ((1 << n) - 1); // XOR the two sets let xor = (set1 ^ set2); // Put the xor bits back to // their original positions xor = (xor << p1) | (xor << p2); // XOR the 'xor' with the original number // so that the two sets are swapped let result = x ^ xor; return result; } // Driver Code let res = swapBits(28, 0, 3, 2); document.write( "Result = " + res); </script> |
输出：
Result = 7
下面是相同逻辑的简短实现
C
int swapBits(unsigned int x, unsigned int p1, unsigned int p2, unsigned int n) { /* xor contains xor of two sets */ unsigned int xor = ((x >> p1) ^ (x >> p2)) & ((1U << n) - 1); /* To swap two sets, we need to again XOR the xor with original sets */ return x ^ ( (xor << p1) | (xor << p2)); } |
JAVA
static int swapBits( int x, int p1, int p2, int n) { /* xor contains xor of two sets */ int xor = ((x >> p1) ^ (x >> p2)) & ((1U << n) - 1 ); /* To swap two sets, we need to again XOR the xor with * original sets */ return x ^ ((xor << p1) | (xor << p2)); } // This code is contributed by subhammahato348. |
C#
static int swapBits( int x, int p1, int p2, int n) { /* xor contains xor of two sets */ int xor = ((x >> p1) ^ (x >> p2)) & ((1U << n) - 1); /* To swap two sets, we need to again XOR the xor with * original sets */ return x ^ ((xor << p1) | (xor << p2)); } // This code is contributed by subhammahato348. |
方法2- 该解决方案的重点是使用与门计算要交换的位的值。然后我们可以根据这些位是否要交换来设置/取消设置这些位。对于要交换的位数（n）——
- 计算移位1=将p1位置的位设置为1后的值
- 计算移位2=将p2位置的位设置为1后的值
- value1=检查位置p1处的num是否已设置的编号。
- value2=检查位置p2处的num是否已设置的数字。
- 如果value1和value2不一样，我们就必须交换位。
例子：
[28 0 3 2] num=28 (11100) p1=0 p2=3 n=2 Given = 11100 Required output = 00111 i.e. (00)1(11) msb 2 bits replaced with lsb 2 bitsn=2 p1=0, p2=3 shift1= 1, shift2= 1000 value1= 0, value2= 1000 After swap num= 10101n=3 p1=1, p2=4 shift1= 10, shift2= 10000 value1= 0, value2= 10000 After swap num= 00111
实施
C++
#include <iostream> using namespace std; int swapBits(unsigned int num, unsigned int p1, unsigned int p2, unsigned int n) { int shift1, shift2, value1, value2; while (n--) { // Setting bit at p1 position to 1 shift1 = 1 << p1; // Setting bit at p2 position to 1 shift2 = 1 << p2; // value1 and value2 will have 0 if num // at the respective positions - p1 and p2 is 0. value1 = ((num & shift1)); value2 = ((num & shift2)); // check if value1 and value2 are different // i.e. at one position bit is set and other it is not if ((!value1 && value2) || (!value2 && value1)) { // if bit at p1 position is set if (value1) { // unset bit at p1 position num = num & (~shift1); // set bit at p2 position num = num | shift2; } // if bit at p2 position is set else { // set bit at p2 position num = num & (~shift2); // unset bit at p2 position num = num | shift1; } } p1++; p2++; } // return final result return num; } /* Driver code*/ int main() { int res = swapBits(28, 0, 3, 2); cout << "Result = " << res; return 0; } |
JAVA
class GFG { static int swapBits( int num, int p1, int p2, int n) { int shift1, shift2, value1, value2; while (n-- > 0 ) { // Setting bit at p1 position to 1 shift1 = 1 << p1; // Setting bit at p2 position to 1 shift2 = 1 << p2; // value1 and value2 will have 0 if num // at the respective positions - p1 and p2 is 0. value1 = ((num & shift1)); value2 = ((num & shift2)); // check if value1 and value2 are different // i.e. at one position bit is set and other it is not if ((value1 == 0 && value2 != 0 ) || (value2 == 0 && value1 != 0 )) { // if bit at p1 position is set if (value1 != 0 ) { // unset bit at p1 position num = num & (~shift1); // set bit at p2 position num = num | shift2; } // if bit at p2 position is set else { // set bit at p2 position num = num & (~shift2); // unset bit at p2 position num = num | shift1; } } p1++; p2++; } // return final result return num; } // Driver code public static void main(String[] args) { int res = swapBits( 28 , 0 , 3 , 2 ); System.out.println( "Result = " + res); } } // This code is contributed by divyeshrabadiya07 |
Python3
def swapBits(num, p1, p2, n): shift1 = 0 shift2 = 0 value1 = 0 value2 = 0 while (n > 0 ): # Setting bit at p1 position to 1 shift1 = 1 << p1 # Setting bit at p2 position to 1 shift2 = 1 << p2 # value1 and value2 will have 0 if num # at the respective positions - p1 and p2 is 0. value1 = ((num & shift1)) value2 = ((num & shift2)) # check if value1 and value2 are different # i.e. at one position bit is set and other it is not if ((value1 = = 0 and value2 ! = 0 ) or (value2 = = 0 and value1 ! = 0 )): # if bit at p1 position is set if (value1 ! = 0 ): # unset bit at p1 position num = num & (~shift1) # set bit at p2 position num = num | shift2 # if bit at p2 position is set else : # set bit at p2 position num = num & (~shift2) # unset bit at p2 position num = num | shift1 p1 + = 1 p2 + = 1 n - = 1 # return final result return num # Driver code res = swapBits( 28 , 0 , 3 , 2 ) print ( "Result =" , res) # This code is contributed by avanitrachhadiya2155 |
C#
using System; class GFG { static int swapBits( int num, int p1, int p2, int n) { int shift1, shift2, value1, value2; while (n-- > 0) { // Setting bit at p1 position to 1 shift1 = 1 << p1; // Setting bit at p2 position to 1 shift2 = 1 << p2; // value1 and value2 will have 0 if num // at the respective positions - p1 and p2 is 0. value1 = ((num & shift1)); value2 = ((num & shift2)); // check if value1 and value2 are different // i.e. at one position bit is set and other it is not if ((value1 == 0 && value2 != 0) || (value2 == 0 && value1 != 0)) { // if bit at p1 position is set if (value1 != 0) { // unset bit at p1 position num = num & (~shift1); // set bit at p2 position num = num | shift2; } // if bit at p2 position is set else { // set bit at p2 position num = num & (~shift2); // unset bit at p2 position num = num | shift1; } } p1++; p2++; } // return final result return num; } // Driver code static void Main() { int res = swapBits(28, 0, 3, 2); Console.WriteLine( "Result = " + res); } } // This code is contributed by divyesh072019 |
Javascript
<script> function swapBits(num, p1, p2, n) { let shift1, shift2, value1, value2; while (n-- > 0) { // Setting bit at p1 position to 1 shift1 = 1 << p1; // Setting bit at p2 position to 1 shift2 = 1 << p2; // value1 and value2 will have 0 if num // at the respective positions - p1 and p2 is 0. value1 = ((num & shift1)); value2 = ((num & shift2)); // Check if value1 and value2 are different // i.e. at one position bit is set and // other it is not if ((value1 == 0 && value2 != 0) || (value2 == 0 && value1 != 0)) { // If bit at p1 position is set if (value1 != 0) { // Unset bit at p1 position num = num & (~shift1); // Set bit at p2 position num = num | shift2; } // If bit at p2 position is set else { // Set bit at p2 position num = num & (~shift2); // Unset bit at p2 position num = num | shift1; } } p1++; p2++; } // Return final result return num; } // Driver code let res = swapBits(28, 0, 3, 2); document.write( "Result = " + res); // This code is contributed by suresh07 </script> |
输出：
Result = 7
参考资料： 用异或交换单个位 如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。
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