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].

翻译：简单的说，就是在给定的数组中寻找元素相减等于给定值k的元素组对数。要求相同的元素对只能用1次。

代码如下：

class Solution{public:int findPairs(vector<int>& nums,int k){if(nums.size() < 2)return 0;if(k< 0)return 0;sort(nums.begin(),nums.end());int left=0;int right=1;int ans=0;while(right<nums.size()){if(left==right)right++;int diff = nums[right]-nums[left];if(diff==k){ans++;num_right=nums[right];right+;while(nums[right]==num_right && right < nums.size())right++;}else if(diff<k){int num_right = nums[right];right++;while(nums[right]==num_right && right < nums.size())right++;}else{int num_left = nums[left];left++;while(nums[left]==num_left && left < right)left++;}}return ans;}};