leetcode笔记：Reverse Bits

一. 题目描述

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).

二. 题目分析

题目的要求比较简单，输入一个32位的无符号整数，根据其二进制表示，输出与其二进制相对称的无符号整数。题目也给出了一个例子。

该题使用基本的位运算即可解决，当然网上也提出了一种很巧妙的方法，其中对于位运算有这样的一种方法，将数字的位按照整块整块的翻转，例如32位分成两块16位的数字，16位分成两个8位进行翻转，以此类推。

对于一个8bit数字abcdefgh来说，其处理过程如下：

abcdefgh -> efghabcd -> ghefcdab -> hgfedcba

进一步的论述，抄送网上的解释：

Remember how merge sort works? Let us use an example of n == 8 (one byte) to see how this works:

        01101001       /        \   0110          1001  /    \        /    \ 01     10     10     01 /\     /\     /\      /\0  1   1  0   1  0     0 1

The first step is to swap all odd and even bits. After that swap consecutive pairs of bits, and so on…

Therefore, only a total of log(n) operations are necessary.

The below code shows a specific case where n == 32, but it could be easily adapted to larger n‘s as well.

三. 示例代码

class Solution {public:    uint32_t reverseBits(uint32_t n) {        uint32_t result = 0;        if (n == result) return result;        int index = 31; // 初始时n的最低位需要右移31位到最高位        while (n)        {            result |= (n & 0x1) << index; // 取n的最低位，并右移到高位            --index; // 右移位数，保持对称            n >>= 1;         }        return result;    }};

// 另一种巧妙的做法/*0x55555555 = 0101 0101 0101 0101 0101 0101 0101 01010xAAAAAAAA = 1010 1010 1010 1010 1010 1010 1010 10100x33333333 = 0011 0011 0011 0011 0011 0011 0011 00110xCCCCCCCC = 1100 1100 1100 1100 1100 1100 1100 1100*/class Solution {public:    uint32_t reverseBits(uint32_t n) {        uint32_t x = n;        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);        return x;    }};

四. 小结

实现该题的要求不难，但是精彩的做法让人大开眼界。