java反转整数中的所有位并返回整数

Question

当我们反转整数1的所有位时，应该返回哪个整数？我们如何用Java代码做到这一点

不应使用java内置函数。不应使用字符串反转、转换为字符串等。只允许按位操作

import java.util.*;
import java.lang.*;
import java.io.*;

class BitReverseInt
{
    public static void main (String[] args) throws java.lang.Exception{
        System.out.println(reverser(1));
    }

    public static int reverser(int given){
          int input = given;
          int temp = 0;
          int output = 0;
          while(input > 0){
            output = output << 1;
            temp = input & 1;
            input = input >> 1;
            output = output | temp;
          }

          return output;
    }
}

Answer 1

位反转可以通过交换相邻的单个位，然后交换相邻的2位字段，然后交换4位字段，依此类推，如下所示。这五条赋值语句可以按任意顺序执行

/********************************************************
 * These are the bit masks used in the bit reversal process

   0x55555555 = 01010101010101010101010101010101
   0xAAAAAAAA = 10101010101010101010101010101010
   0x33333333 = 00110011001100110011001100110011
   0xCCCCCCCC = 11001100110011001100110011001100
   0x0F0F0F0F = 00001111000011110000111100001111
   0xF0F0F0F0 = 11110000111100001111000011110000
   0x00FF00FF = 00000000111111110000000011111111
   0xFF00FF00 = 11111111000000001111111100000000
   0x0000FFFF = 00000000000000001111111111111111
   0xFFFF0000 = 11111111111111110000000000000000

 */

    uint x = 23885963;    // 00000001011011000111100010001011

    x = (x & 0x55555555) <<  1 | (x & 0xAAAAAAAA) >>  1; 
    x = (x & 0x33333333) <<  2 | (x & 0xCCCCCCCC) >>  2; 
    x = (x & 0x0F0F0F0F) <<  4 | (x & 0xF0F0F0F0) >>  4; 
    x = (x & 0x00FF00FF) <<  8 | (x & 0xFF00FF00) >>  8; 
    x = (x & 0x0000FFFF) << 16 | (x & 0xFFFF0000) >> 16;

    // result x == 3508418176   11010001000111100011011010000000

通过查看每个中间结果，你可以看到正在发生什么

希望这能给你所需要的，在你的头脑中解决它。John Doe的答案将第4步和第5步合并为一个表达式。这在大多数机器上都适用

Answer 2

在Java中，有多种方法可以反转给定数字的位

首先，Java语言有内置的位补码运算符（~）。所以（~number）反转数字的位

其次，可以使用整数。反向（数字）

第三，如果这是测试的一部分，或者你只是想玩比特，你可以参考下面的代码

/*
The logic uses moving bitmask from right to left:

 1. Get the bit of given number, by binary and(&) with bitmask
 2. XOR(^) with the bitmask, so here we reverse the bit.
 3. OR(|) this reversed bit with the result(result has all Zero initially)

This logic is repeated for each 32 bits by moving the mask from right to left, 
one bit at a time using (<<) left shift operator on bitmask.
*/
public class ReverseBits {

    public static int reverseBits(int input) {
        print("Input", input);
        int bitmask = 1;
        int result = 0;
        do {
            //print("Bitmask", bitmask);
            result = result | (bitmask ^ (input & bitmask)) ;
            //print("Result", result);
            bitmask = bitmask << 1;
        } while (bitmask != 0);
        print("Reverse", result);
        return result;
    }

    public static void print(String label, int input) {
        System.out.println(label +"\t:"+Integer.toBinaryString(input));
    }

    public static void main(String[] args) {
        reverseBits(Integer.MIN_VALUE);
        reverseBits(Integer.MAX_VALUE);
        reverseBits(reverseBits(170));
    }
}

输出：

Input   :10000000000000000000000000000000
Reverse :1111111111111111111111111111111
Input   :1111111111111111111111111111111
Reverse :10000000000000000000000000000000
Input   :10101010
Reverse :11111111111111111111111101010101
Input   :11111111111111111111111101010101
Reverse :10101010