## 题目描述：

LeetCode 532. 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].

## 题目大意：

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：

```首先将nums中的数字放入字典c

## Python代码：

``````class Solution(object):
def findPairs(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
if k < 0: return 0
c = collections.Counter(nums)
return sum(c[n + k] > 1 - bool(k) for n in c.keys())
``````

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1. 书影网友 发布于 2017年9月18日 11:31 #

有点看不懂你的思路，代码
```
c[n + k] &amp;gt; 1 - bool(k)
```

是什么意思？