K-diff Pairs in an Array

Question

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.

Solution

可以用非常簡單的 Set 解決，但總覺得很難想（連看解答我也覺得一知半解）。還是覺得用 Map 比較直觀

public int findPairs(int[] nums, int k) {
        if (k < 0) { return 0; }

        Set<Integer> starters = new HashSet<Integer>();
        Set<Integer> uniqs = new HashSet<Integer>();
        for (int i = 0; i < nums.length; i++) {
            if (uniqs.contains(nums[i] - k)) starters.add(nums[i] - k);
            if (uniqs.contains(nums[i] + k)) starters.add(nums[i]);
            uniqs.add(nums[i]);
        }

        return starters.size();
    }

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;
    }