题目地址：https://leetcode.com/problems/reverse-bits/
题目分析：可以4bit为单位，0翻转对应0，1翻转对应8.....15翻转对应15，将这些翻转信息保存在数组中即可以O(1)的空间复杂度换来很好的时间复杂度
题目解答：
public class Solution { // you need treat n as an unsigned value public int reverseBits(int n) { int[] tb = {0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15}; int ret = 0; int msk = 15; for(int i = 0;i<8;i++){ int curr = n&msk; ret = ret<<4; ret |= tb[curr]; n = n>>>4; } return ret; } }
原文：http://www.cnblogs.com/xiongyuesen/p/4417990.html