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].

思路：

用一个map记录所有关键字，这样还能同时去掉了重复元素，然后遍历map，统计满足条件个数

注意 absolute difference 绝对值>=0

int findPairs(vector
     
      & nums, int k)    {        if (k < 0)return 0;        int result = 0;        //sort(nums.begin(),nums.end());//不需要排序        map
      
       table;        map
       
        ::iterator it;        for (int i = 0; i < nums.size();i++)        {            table[nums[i]]++;        }        for (it = table.begin(); it != table.end();it++)        {            if (k!=0 &&table.count(k + it->first))//比如此时first为1 而k=2 判断是否存在3            {                result++;            }            else if (k == 0 && it->second >= 2)//说明存在至少两个元素值相等            {                result++;            }        }        return result;    }