[LeetCode] 477. Total Hamming Distance（位操作）

最新推荐文章于 2025-05-22 01:15:00 发布

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最新推荐文章于 2025-05-22 01:15:00 发布

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原文链接：http://www.cnblogs.com/ZhaoxiCheung/p/7400361.html

本文介绍了一种计算数组中所有数对汉明距离之和的方法。通过统计每一位上1和0的数量，利用这些数量计算出所有配对的汉明距离总和。

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Description

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

题意：给定一个数组，其中两两配对，求出所有配对数的汉明距离。

题解：我们知道，汉明距离是两个数其二进制位上不同的位数，因此，对于32位数，遍历二进制位，统计每个数在第一个bit位有多少个数是1，有多少个数是0，然后统计第二个bit位，以此类推。

class Solution {
public:
    //59ms
    int totalHammingDistance(vector<int>& nums) {
        int res = 0,len = nums.size();
        for (int i = 0;i < 32;i++){
            int cnt = 0;
            for (int j = 0;j < len;j++){
                cnt += (nums[j] >> j) & 1;
            }
            res += (len - cnt) * cnt;
        }
        return res;
    }
};

转载于:https://www.cnblogs.com/ZhaoxiCheung/p/7400361.html