477. Total Hamming Distance

标签：different   暴力破解   int   get   gdi   tween   ++   问题   put   

问题描述：

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Now your job is to find the total Hamming distance between all pairs of the given numbers.

Example:

Input: 4, 14, 2

Output: 6

Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
showing the four bits relevant in this case). So the answer will be:
HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.

Note:

1. Elements of the given array are in the range of 0 to 10^9
2. Length of the array will not exceed 10^4.

解题思路：

一开始也是首先想到了暴力破解法，不过我觉得应该是过不了OJ的

看了GrandYang的解法

对所有数字的每一位统计1的个数和0的个数

这一位的总 汉明距离 为1的个数乘以0的个数

代码：

class Solution {
public:
    int totalHammingDistance(vector<int>& nums) {
        int ret = 0;
        int n = nums.size();
        for(int i = 0; i < 32; i++){
            int zero = 0;
            int one = 0;
            int mode = 1;
            mode <<= i;
            for(int j = 0; j < nums.size(); j++){
                if(nums[j] & mode) one++;
                else zero++;
            }
            ret += zero * one;
        }
        return ret;
    }
};

477. Total Hamming Distance

标签：different   暴力破解   int   get   gdi   tween   ++   问题   put   

原文地址：https://www.cnblogs.com/yaoyudadudu/p/9276202.html