Leetcode 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].

刚看到这个问题的时候，很自然地想到一个遍历搞定了，但是实际上是有问题的。因为题目要求不重复，且与顺序无关。

那么就想到用hashmap了。

但是要怎么放进去一对数字呢。

其实不然，差是固定的，要比较的数也是呀。为什么一定要放一对数进去而不放单个数进去再寻找呢。

其实是这个意思

public class Solution {    public int findPairs(int[] nums, int k) {        if (nums == null || nums.length == 0 || k < 0)   return 0;//说了一万遍 要检查null和0！！！                Map<Integer, Integer> map = new HashMap<>();        int count = 0;        for (int i : nums) {            map.put(i, map.getOrDefault(i, 0) + 1);//把数依次放进去        }                for (Map.Entry<Integer, Integer> entry : map.entrySet()) {            if (k == 0) {                               if (entry.getValue() >= 2) {//统计重复出现的数字个数                    count++;                }             } else {//k不为0的时候

               if (map.containsKey(entry.getKey() + k)) {                    count++;                }            }        }                return count;    }}