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.
1，首先想到hashSet去重复，但是有k=0的情况，所以不适用
2，用hashMap，k=0，value>=2，避免反复出现[i,i,i,i]的情况
public class Solution { public int findPairs(int[] nums, int k) { if(nums.length == 0 ||nums == null || k < 0) return 0; int m = 0; Map<Integer, Integer> map = new HashMap<>(); for(int c: nums) { if(map.containsKey(c)) { map.put(c, map.get(c) + 1); } else { map.put(c, 1); } } for(Map.Entry<Integer, Integer> entry: map.entrySet()) { if(k == 0) { if(entry.getValue() >= 2) { m++; } } else { if(map.containsKey(entry.getKey() + k)) { m++; } } } return m; } }1，首先想到hashSet去重复，但是有k=0的情况，所以不适用
2，用hashMap，k=0，value>=2，避免反复出现[i,i,i,i]的情况
