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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
题目：把一个32位的unsigned integer 翻转过来
可以一位一位的移，也可以更多位的移，建立一个hash表即可，我选择的是4位4位的移
class Solution { public: uint32_t reverseBits(uint32_t n) {
//ret用来保存返回值，assist用来辅助 uint32_t ret = 0,assist=0;
//建立map，用来保存0~15对应的翻转无符号数 map<uint32_t, uint32_t> ma; ma[0] = 0;ma[1] = 8;ma[2] = 4;ma[3] = 12;ma[4] = 2;ma[5] = 10;ma[6] = 6;ma[7] = 14; ma[8] = 1;ma[9] = 9;ma[10] = 5;ma[11] = 13;ma[12] = 3;ma[13] = 11;ma[14] = 7;ma[15] = 15;
//每次移4位，移7次即可 for (int i = 0;i < 7;++i) {
//用来得到前4位对应的翻转数 assist = ma[n & 15ul]; ret += assist; n=n >> 4; ret=ret << 4; } assist = ma[n & 15ul]; ret += assist; return ret; } };
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原文地址：http://www.cnblogs.com/csudanli/p/5353627.html
