532. K-diff Pairs in an Array

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        unordered_map<int, int> exist;
        int res = 0;
        if (k < 0) return 0;    //因为是absolute diff（觉得纯粹深井冰
        for (int num : nums) {
            if (exist.count(num)) {
                exist[num]++;
                if (!k && exist[num] == 2) res++;
                continue;
            }    
            //考虑重复的数字，考虑数字重复时k=0怎么办，啊 然而[1,1,1,1,1]，0 如果用一个set没辙
            if (exist.count(num + k)) res++;
            if (k && exist.count(num - k)) res++;
            exist[num]++;
        }
        return res;
    }
};

更干净的方法：直接count然后一次过计算

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;
    }
}