LeetCode刷题【Array】 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].

解决方法一： 每一个数与它距离为k的数最多两个;为了去除重复对, 数+k,或数-k,考虑到重复情况,只需要考虑数+k在map中找是否存在对应数; Runtime: 37 ms 

public class Solution {    public int findPairs(int[] nums, int k) {        if (nums == null || nums.length == 0 || k < 0)   return 0;                Map<Integer, Integer> map = new HashMap<>();        int count = 0;        for (int i : nums) {            map.put(i, map.getOrDefault(i, 0) + 1);        }                for (Map.Entry<Integer, Integer> entry : map.entrySet()) {            if (k == 0) {                //count how many elements in the array that appear more than twice.                if (entry.getValue() >= 2) {                    count++;                }             } else {                if (map.containsKey(entry.getKey() + k)) {                    count++;                }            }        }                return count;    }}

参考：

【1】https://leetcode.com/