190. Reverse Bits

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

链接： http://leetcode.com/problems/reverse-bits/

2/21/2017, Java

参考别人的算法

错误：

第7行百思不得其解，为什么res = res + (n >> i) & 1是错的，而res += (n >> i) & 1是对的。原因是+/-的优先级居然比位运算要高！

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int res = 0;
        for (int i = 0; i < 32; i++) {
            res = res << 1;
            res += (n >> i) & 1;
        }
        return res;
    }
}

据说还有其他算法：

Bit Twiddling Hacks以及Hacker's Delight

准备留给二刷