190. Reverse Bits

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?

Related problem: Reverse Integer.

Solution1:

刚开始没什么想法，看similar problem是1bit那道，大概有个了个思路。从least significant判断是不是1，每读一次，向右移一位，然后用一个0buffer xor（加上）这个least signifciant的值。

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int res=0;
        for(int i=0;i<32;i++)
        {
            if(((n>>i)&1)==1)
            {
                res^=(1<<(31-i));
            }
        }
        return res;

    }
}

190. Reverse Bits

原文：http://www.cnblogs.com/Machelsky/p/5871318.html