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 = 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:[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: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: 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].

讲道理跟Two Nums很像，可惜没出方法，用了个比较土的方法，第二个参考的方法会贴上类似Two Nums的方法

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {

        if(k < 0)
            return 0;
        
        multiset<int> s(nums.begin(), nums.end());
        //int n = nums.size();
        int count = 0;

        if(k == 0)
        {
            for (auto n : s)
            {
                if (s.count(n) >= 2)
                {
                    ++count;
                    s.erase(n);
                }
            }
        }
        if (k == 0)
            return count;

        set<int> t(nums.begin(), nums.end());
        for (auto n : t)
        {
            int target = n + k;
                
            if (t.find(target) != t.end())
            {

                ++count;
            }
        }

        return count;
    }
};

法二：

class Solution {
public:
    /**
     * for every number in the array:
     *  - if there was a number previously k-diff with it, save the smaller to a set;
     *  - and save the value-index to a map;
     */
    int findPairs(vector<int>& nums, int k) {
        if (k < 0) {
            return 0;
        }
        unordered_set<int> starters;
        unordered_map<int, int> indices;
        for (int i = 0; i < nums.size(); i++) {
            if (indices.count(nums[i] - k)) {//target1
                starters.insert(nums[i] - k);
            }
            if (indices.count(nums[i] + k)) {//target2
                starters.insert(nums[i]);
            }

            indices[nums[i]] += 1;
        }
        
        return starters.size();
    }
};