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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 as 00111001011110000010100101000000).
Follow up: If this function is called many times, how would you optimize it?
思路：二进制数反转。
代码：
方法一：
1 class Solution { 2 public: 3 uint32_t reverseBits(uint32_t n) { 4 n = (n >> 16) | (n << 16); 5 n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8); 6 n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4); 7 n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2); 8 n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1); 9 return n; 10 } 11 };
方法二：
1 class Solution { 2 public: 3 uint32_t reverseBits(uint32_t n) { 4 int i; 5 uint32_t res=0; 6 for(i=0;i<32;i++){ 7 res<<=1; 8 res|=n&1; 9 n>>=1; 10 } 11 return res; 12 } 13 };
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原文地址：http://www.cnblogs.com/Deribs4/p/5721344.html
