532. K-diff Pairs in an Array

Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array.

A k-diff pair is an integer pair (nums[i], nums[j]), where the following are true:

• 0 <= i, j < nums.length
• i != j
• |nums[i] - nums[j]| == k

Notice that |val| denotes the absolute value of val.

Example 1:

Input: nums = [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: nums = [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: nums = [1,3,1,5,4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).

Example 4:

Input: nums = [1,2,4,4,3,3,0,9,2,3], k = 3
Output: 2

Example 5:

Input: nums = [-1,-2,-3], k = 1
Output: 2

Constraints:

• 1 <= nums.length <= 104
• -107 <= nums[i] <= 107
• 0 <= k <= 107

class Solution {
    public int findPairs(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap();
        for(int i : nums) {
            map.put(i, map.getOrDefault(i, 0) + 1);
        }
        int res = 0;
        for(int i : nums) {
            if(k == 0 && map.get(i) > 1) {
                res++;
                map.put(i, 0);
            }
            else if(k != 0 && map.containsKey(i + k) && map.get(i + k) != 0) {
                res++;
                map.put(i + k, 0);
            }
        }
        return res;
    }
}

特殊情况：k == 0，先把数字和频率存到map

然后要判断k是否=0，如果是，就看这个数是否出现2次或以上，然后把频率变0（防止重复）

k≠0，就看有没有对应的key，有就更新频率为0（防止重复）