leetcode 532. K-diff Pairs in an Array_数组中的k-diff数对归并排序-CSDN博客

本文链接：https://blog.csdn.net/liqinzhe11/article/details/79023857

本文介绍了一种算法解决方案，用于在整数数组中找出所有唯一k-diff数对的数量。通过使用C++中的map数据结构，该算法能有效地解决这一问题，并提供了几个示例来说明其工作原理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

532. K-diff Pairs in an Array

Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.

Example 1:

Input: [3, 1, 4, 1, 5], k = 2
Output: 2
Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number of unique pairs.

Example 2:

Input:[1, 2, 3, 4, 5], k = 1
Output: 4
Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).

Example 3:

Input: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).

Note:

The pairs (i, j) and (j, i) count as the same pair.
The length of the array won't exceed 10,000.
All the integers in the given input belong to the range: [-1e7, 1e7].

1、绝对值不可能为负数。

2、k有可能为0，所以要先 it->second --;

3、map必须先判断在不在里面，再判断value是否大于0。不能直接判断，否则map会自动创建。

class Solution {
public:
    int findPairs(vector<int>& nums, int k)
    {
        if (k < 0) return 0;
        map<int, int> mp;
        for (auto it : nums)
            mp[it] ++;
        int ret = 0;
        for (auto it = mp.begin(); it != mp.end(); it++)
        {
            it->second --;
            if (mp.find(it->first + k) != mp.end() && mp[it->first + k] > 0)
                ret ++;
        }
        return ret;
    }
};