[leetcode]477. Total Hamming Distance

最新推荐文章于 2025-05-22 01:15:00 发布

xiaocong1990

最新推荐文章于 2025-05-22 01:15:00 发布

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本文介绍了一种高效算法，用于计算一组整数中所有数对之间的总汉明距离。通过统计每个比特位上的1出现次数并计算贡献值，实现O(n)时间复杂度的解决方案。

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题目链接:https://leetcode.com/problems/total-hamming-distance/

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.

解题思路：数组大小为size，统计每个数在同一个bit位的出现1的个数，记为bitCount,那么这个bit位贡献的汉明距离为bitCount*(size-bitCount)。总的时间复杂度为O(size)。

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