532. K-diff Pairs in an Array

题目：

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].

思路：

本题思路如下，分为如下几个步骤：

step1:先对数组进行排序

step2:查找当前数与k的差值是否在数组中存在（当前位置的数字除外）

step3:剔除重复项

代码：

class Solution {public:    int findPairs(vector<int>& nums, int k) {        sort(nums.begin(),nums.end());        int count = 0;        vector<pair<int,int>> res;        if(k<0)            return 0;        for(int i =nums.size()-1;i>0;i--)        {            if(find(nums.begin(),nums.end(),nums[i]-k)!=nums.end()&&find(nums.begin(),nums.end(),nums[i]-k)!=nums.begin()+i)            {                res.push_back(make_pair(nums[i],nums[i]-k));                nums.erase(nums.begin()+i);                count++;            }        }        for(int j=1;j<res.size();j++)        {            if(res[j].first==res[j-1].first)                count--;        }        return count;     }};