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
public class Solution { // you need treat n as an unsigned value public int reverseBits(int n) { //注意每个位的下标关系，新的位数+旧的位数=31 int res=0; for(int i=0;i<32;i++){ /* res=res<<1; res|=n&1; n=n>>1;*/ res|=(((n>>i)&1)<<(31-i)); } return res; } }
原文：http://www.cnblogs.com/qiaomu/p/4700297.html