LeetCode 532. K-diff Pairs in an Array

K-diff Pairs in an Array

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题目描述:

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.

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].

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题目大意:

给定一个数组nums和一个数字k，找出相差k的两个数，称这两个数位一对，并且每一对数组顺序不一样算同一种情况，求数组中有多少对。
我们只需要遍历每一个数字，并且对于每一个数字，都判断这个数字后面是否有和他相差k的数字即可。
重点在于如何去重，我们可以把他排序，排序后相同的元素相邻，去重变得容易，而且我们可以在对每一个数字判断是否有和他相差k的数字的时候，只需要考虑他后面比他大的数字即可。

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题目代码:

class Solution {public:    int findPairs(vector<int>& nums, int k) {        if(nums.size() == 0) return 0;        sort(nums.begin(), nums.end());        int cnt = 0;        int i = 0;        while(i < nums.size()-1){            for(int j = i+1; j < nums.size(); j++){                if(nums[i]+k == nums[j]){                    cnt++;                    break;                }                else if(nums[i]+k < nums[j])                    break;            }            i++;            while(nums[i] == nums[i-1]) i++;        }            return cnt;    }};