LeetCode Total Hamming Distance

最新推荐文章于 2025-08-13 10:25:05 发布

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最新推荐文章于 2025-08-13 10:25:05 发布

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文章标签：数据结构与算法 java

原文链接：http://www.cnblogs.com/Dylan-Java-NYC/p/6242014.html

本文讲解了如何计算一组整数中所有数对之间的总Hamming距离，并提供了一种高效的算法实现。通过遍历int类型的32位，计算每位上1的数量，进而得出总的Hamming距离。

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原题链接在这里：https://leetcode.com/problems/total-hamming-distance/

题目：

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.

题解：

Hamming Distance进阶题目.

对于int的32位的每一位，计算在这一位是1的array element个数k, 那么在这一位是0的个数就是nums.length-k. 对于这一位就有k*(nums.length-k)种选法可以产生Hamming Distance. 然后将32位Hamming Distance 个数相加就是总数.

Time Complexity: O(1). Space: O(1).

AC Java:

 1 public class Solution {
 2     public int totalHammingDistance(int[] nums) {
 3         int res = 0;
 4         int len = nums.length;
 5         
 6         for(int i = 0; i< 32; i++){
 7             int bitCount = 0;
 8             for(int j = 0; j<len; j++){
 9                 bitCount += (nums[j] >> i) & 1;
10             }
11             res += bitCount*(len - bitCount);
12         }
13         
14         return res;
15     }
16 }