K-diff Pairs in an Array问题及解法

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 = 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.

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).

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].

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();    }};