[LeetCode] 190. Reverse Bits

https://leetcode.com/problems/reverse-bits/

最简单的想法是位运算一位一位的挪，用for循环搞定（要注意优先级和结合性的问题）：

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

view raw reverseBits.java hosted with ❤ by GitHub

然后看到了别人一个不用循环的解法：
https://leetcode.com/problems/reverse-bits/discuss/54741/O(1)-bit-operation-C++-solution-(8ms)
很巧妙的利用了交换的时候这种特性：abcdefgh -> efghabcd -> ghefcdab -> hgfedcba 好困啊，只好用简单题交差啦 ～(￣▽￣～)(～￣▽￣)～