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 = 2Output: 2Explanation: 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 = 1Output: 4Explanation: 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 = 0Output: 1Explanation: There is one 0-diff pair in the array, (1, 1).

Note:

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

找数组里绝对差值为k的pairs。这道题比较tricky的一点就是当k=0的时候不好处理。因此可以用一个unordered_map保存数组中每个数出现的次数，然后对k=0以及k!=0分别处理。

class Solution {public:    int findPairs(vector<int>& nums, int k) {        unordered_map<int, int> res;        for(auto num : nums)          res[num]++;        int count = 0;        for(auto iter : res){            if(k == 0){                if(iter.second >= 2) count++;            }            else{                if(res.find(iter.first + k) != res.end()) count++;            }        }                return k < 0 ? 0 : count;    }};