首页 > 其他 > 详细

477. Total Hamming Distance

时间:2020-04-13 12:23:37      阅读:47      评论:0      收藏:0      [点我收藏+]

Problem:

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:

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

思路

Solution (C++):

int totalHammingDistance(vector<int>& nums) {
    int n = nums.size(), res = 0;
    if (n < 2)  return 0;
    vector<int> diff(2, 0);
    
    while (true) {
        int count = 0;
        diff[0] = 0;
        diff[1] = 0;
        
        for (int i = 0; i < n; ++i) {
            if (nums[i] == 0)  ++count;
            ++diff[nums[i]%2];
            nums[i] >>= 1;
        }
        res += (diff[0] * diff[1]);
        
        if (count == n)  return res;
    }
}

性能

Runtime: 80 ms??Memory Usage: 8.2 MB

思路

Solution (C++):


性能

Runtime: ms??Memory Usage: MB

477. Total Hamming Distance

原文:https://www.cnblogs.com/dysjtu1995/p/12690411.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!