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:
The pairs (i, j) and (j, i) count as the same pair.
The length of the array won't exceed 10,000.
All the integers in the given input belong to the range: [-1e7, 1e7].

这道题目，如果使用c++ set unordered_map 的话，有如下解法：

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        set<pair<int,int>> unq;
        unordered_map<int,int> m;
        
        sort(nums.begin(),nums.end());
        
        for(int i = 0;i < nums.size();i++)
        {
            int target = nums[i] - k;
            if(m.find(target) != m.end())
            {
                unq.insert(make_pair(nums[i],target));
            }
            m[nums[i]] = 1; 
        }
        
        return unq.size();
     }
};

如果你不理解for循环中最后赋值语句的作用（这里赋任何值都ok，作用是使上面的find函数条件成立)，可以编写demo验证{3,1,4,1,5} 2

#include <iostream>
#include <set>
#include <unordered_map>
#include <vector>
#include <algorithm>

using namespace std;

int findPairs(vector<int>& nums, int k) 
{
    set<pair<int,int>> unq;
    unordered_map<int,int> m;
    
    sort(nums.begin(),nums.end());
    
    for(int i = 0;i < nums.size();i++)
    {
        int target = nums[i] - k;
        if(m.find(target) != m.end())
        {
            unq.insert(make_pair(nums[i],target));
        }
        m[nums[i]] = 1; 
    }
    
    return unq.size();
}

int main()
{
    vector<int> nums = {3,1,4,1,5};
    int res = findPairs(nums,2);
    cout << "res is "<< res <<endl;

    return 0;    
}

另外一种效率相对第一种高的解法：

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        int cnt = 0;
        if(k < 0)
            return cnt;
        
        unordered_map<int,int> m;
     
        for(auto it = nums.begin();it != nums.end();it++)
            m[*it]++;
        
        for(auto mit = m.begin();mit != m.end();mit ++)
        {
            if (0 != k)
            {
                if (m.find(mit->first + k) != m.end())
                    cnt++;
            }else
            {
                //统计重复元素个数,例如[1,1,2,2,3,3] 0,second >1的有1 2 3
                if(mit->second > 1)
                    cnt++;
            }
        }
       return cnt;
    }
};