LeetCode OJ 之 Reverse Bits （翻转二进位）

题目：

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as00111001011110000010100101000000).

Follow up:
If this function is called many times, how would you optimize it?

Related problem: Reverse Integer

翻转一个数字的二进制位。

思路：

从后依次向前求出第i位，然后左移 31-i 位，相加即可。

代码：

uint32_t reverseBits(uint32_t n) {    uint32_t result = 0;    for(int i = 0 ; i < 32 ; i++)    {        uint32_t tmp = ((n >> i) & 1);//从右向左依次求出 n 的每一位        tmp = tmp << (31 - i);//左移31 - i 位，注意左移并不改变原来的tmp，必须把左移后的结果再赋给tmp        result += tmp;//求和    }    return result;}