题目：
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：
该题的思路与Reverse Integer类似,这里是对二进制数进行转置，我们可以使用移位的方法进行，对原数不断取最后一位：n & 1
然后不断右移：n=n>>1
而对结果数不断左移或上原数的最后一位：res=res<<1; res=res | (n & 1)
由于位数是确定的，因此只需要移位31次即可。
代码如下：
public class Solution { // you need treat n as an unsigned value public int reverseBits(int n) { int res= n & 1; for(int i=1;i<=31;i++) { n=n>>1; //不断向右移动 res=res<<1; //不断向左移动 res=res | (n & 1); } return res; } }
reference: http://pisxw.com/algorithm/Reverse-Bits.html
转载于:https://www.cnblogs.com/hygeia/p/4687757.html
相关资源：JAVA上百实例源码以及开源项目