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LeetCode 190: Reverse Bits

时间:2015-03-19 17:48:43      阅读:123      评论:0      收藏:0      [点我收藏+]

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题目描述:

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?

思路:

主要考察位运算

1.result=0,result=result + 将数字和1做位与运算,对数字不停做位右移操作,直至数字为0;

2.将32位数字补全

    uint32_t reverseBits(uint32_t n) {
        uint32_t result;
        int num = 0;
        while(n){
            result = result *2 + (uint32_t)(n&1);
            n = n>>1;
            num++;
        }
        while(num<32){
            result = result<<1;
            num++;
        }
        return result;
    }

  

LeetCode 190: Reverse Bits

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原文地址:http://www.cnblogs.com/xiamaogeng/p/4350824.html

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