LeetCode 190 :Reverse Bits

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).

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

Related problem: Reverse Integer

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

分析：

题目要求将一个整数的二进制翻转，如0x00000001 翻转成0x80000000。这个与10进的翻转类似。

代码如下：

uint32_t reverseBits(uint32_t n) {
        int result = 0;
        for(int i=0;i<31; i++){
            //取最后一位，然后右移
            int tmp = n & 0x01;
            n = n >> 1;
            //先左移一位，与上最后一位
            result = (result<<1) | tmp;
        }
        return result;
    }

LeetCode 190 :Reverse Bits

原文：http://blog.csdn.net/sunao2002002/article/details/45652165