Leetcode 190. Reverse Bits

版权声明：博客文章都是作者辛苦整理的，转载请注明出处，谢谢！ https://blog.csdn.net/Quincuntial/article/details/82389859

文章作者：Tyan
博客：noahsnail.com  |  CSDN  |  简书

1. Description

2. Solution

• Version 1

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        int a[32] = {0};
        unsigned int result = 0;
        for(int i = 0; i < 32; i++) {
            a[i] = n & 1;
            n >>= 1;
        }
        unsigned int m = 1;
        for(int i = 31; i >= 0; i--) {
            result += a[i] * m;
            m <<= 1;
        }
        return result;
    }
};

• Version 2

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t x = 2147483648;
        uint32_t result = 0;
        while(n) {
            int temp = n & 1;
            result += temp * x;
            n >>= 1;
            x >>= 1;
        }
        return result;
    }
};

• Version 3

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t result = 0;
        for(int i = 0; i < 32; i++) {
            result <<= 1;
            result |= (n & 1);
            n >>= 1;
        }
        return result;
    }
};

• Version 4

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t result = 0;
        for(int i = 0; i < 32; i++) {
            result <<= 1;
            if(n & 1) {
                result++;
            }
            n >>= 1;
        }
        return result;
    }
};

Reference

1. https://leetcode.com/problems/reverse-bits/description/