问题描述：
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.
示例：
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. 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). 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]. 问题分析：求解的元组（i,j）满足 i =j + k 或者 j = i + k，根据这个性质，我们采用unordered_set 和 unordered_map来求解问题。set用于存储唯一的结果，map用于保证数组中的值的唯一性。
过程详见代码：
class Solution { public: int findPairs(vector<int>& nums, int k) { if (k < 0) { return 0; } unordered_set<int> starters; unordered_map<int, int> indices; for (int i = 0; i < nums.size(); i++) { if (indices.count(nums[i] - k)) { starters.insert(nums[i] - k); } if (indices.count(nums[i] + k)) { starters.insert(nums[i]); } indices[nums[i]] += 1; } return starters.size(); } };