[LeetCode] Reverse Bits

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 as00111001011110000010100101000000).

Follow up:
If this function is called many times, how would you optimize it?

解题思路：

通过移位来计算。下面是代码：

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t result=0;
        int size = sizeof(uint32_t)*8;
        for(int i=0; i<size; i++){
            result = result << 1;
            result += n%2;
            n=n>>1;
        }
        return result;
    }
};

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t result=0;
        int size = sizeof(uint32_t)*8;
        int i=0;        //已经移动的位数
        while(n!=0){
            result = result << 1;
            result += n%2;
            n=n>>1;
            i++;
        }
        result <<= size - i;
        return result;
    }
};

[LeetCode] Reverse Bits

原文：http://blog.csdn.net/kangrydotnet/article/details/44920479