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).


解法 :
      用一個map 一個儲存這個array中會有幾個element以及他出現幾次 然後找尋這些元素中有沒有符合 i + k也存在的
不過這邊要特別注意，當k = 0時要特別處理，不然第三種條件下你會變成每個元素通通符合。 處理k = 0的條件是這樣，只要同樣元素的數量大於一就代表也算一個pair。

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        
        int counter = 0;
        unordered_map<int, int> elementmap;
        
        for (int i : nums)
        {
            elementmap[i]++;
        }
        
        for (auto pair : elementmap) 
        {
            if (k == 0 && pair.second > 1) 
            {
                counter++;
            }
            
            if (k > 0 && elementmap.count(pair.first + k))
            {
                counter++;
            }
        }
        
        return counter;
    }
};