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?
int型整数对应二进制数移位操作，代码如下：
public class Solution { // you need treat n as an unsigned value public int reverseBits(int n) { int re = 0; int bit = 0; for(int i=0;i<32;i++) { bit = n&1;//取出n对应二进制数的最低位 re = re|(bit<<31-i); n = n>>1; } return re; } }
当然也可以这样写：
public class Solution { // you need treat n as an unsigned value public int reverseBits(int n) { return Integer.reverse(n); } }
原文：http://www.cnblogs.com/mrpod2g/p/4325713.html