LeetCode-algorithms 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 = 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).

思路：

这题的目的是找绝对距离的，所以首先对数组进行排序，会使得运算方便很多。排序之后只需要对每一个位置i，往后j个数进行遍历，当A[i] - A[j] > k时，便可以开始下一个i的遍历；其次，当j和j+1相同时，也直接跳过。

代码：

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

结果：