# 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](https://en.wikipedia.org/wiki/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].

## 题目大意：

给定一个整数数组nums，以及一个整数k，找出其中所有差恰好为k的不重复数对。

**注意：**

1. 数对(i, j) 和 (j, i)算作同一个数对
2. 数组长度不超过10,000
3. 所有整数在范围\[-1e7, 1e7]之间

## 解题思路：

**字典（Map）**

```
首先将nums中的数字放入字典c

遍历c，记当前数字为i
分为两种情况：
（1）k==0，即找出值相等的对数。

（2）k!=0，若i + k在c中，则将结果+1
```

```
class Solution(object):
    def findPairs(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        res = 0
        c = collections.Counter(nums)
        for i in c:
            if k > 0 and i + k in c or k == 0 and c[i] > 1:
                res += 1
        return res
```
