[leedcode 190] Reverse Bits

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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 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;
    }
}

[leedcode 190] Reverse Bits

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原文地址：http://www.cnblogs.com/qiaomu/p/4700297.html