LeetCode #477 - Total Hamming Distance-优快云博客

本文介绍了一种高效计算一组整数中所有数对间汉明距离总和的方法。通过统计每位上0和1的数量来避免了直接两两比较的时间复杂度问题，实现了O(n)时间复杂度的解决方案。

题目描述：

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:

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

计算一个数组中每对数之间的汉明距离之和。如果直接遍历数组两两求解，时间复杂度为O(n*n)，但是如果我们根据32位，计算在某一位上数组中有x个数是0，有y个数在这一位是1，就可以得到在这一位上汉明距离之和为x*y，也就是x个数和y个数的两两组合，这样的时间复杂度为O(n)。

class Solution {
public:
    int totalHammingDistance(vector<int>& nums) {
        int result=0;
        for(int i=0;i<32;i++)
        {
            int mask=1<<i;
            int count=0;
            for(int j=0;j<nums.size();j++)
                if((mask&nums[j])>0) count++;
            result+=count*(nums.size()-count);
        }
        return result;
    }
};