[LeetCode] 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:

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].

找出数组中元素差值为k的元素对的数量。要求元素对不重复。

思路：首先对数组进行排序，才能利用差为k这个条件顺序遍历，然后使用两个指针对数组遍历。类似于选择排序 ，将差值为k的两个元素的和放入set中。最后遍历结束返回set的大小即可。

之所以使用set是因为题目要求元素对不可以重复。

成功Accept但是Runtime感人。

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        if (nums.empty())
            return 0;
        unordered_set<int> s;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size() - 1; i++) {
            for (int j = i + 1; j < nums.size(); j++) {
                if (nums[j] - nums[i] == k)
                    s.emplace(nums[j] + nums[i]);
            }
        }
        return s.size();
    }
};
// 813 ms