Leetcode_Array -- 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( 绝对差 |x-y| ) 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].

解决方案：

Python

(1)
class Solution(self,nums,k)
    def findpair(self,nums,k):
        uniquelist = set(nums)
        count = 0
        if k<0:
            retrun 0
        elif k == 0:
            for i in uniquelist:
                if nums.count(i)>1:
                    count += 1
            return count
        else k > 0:
            for i in uniquelist:
                if k+i in uniquelist:
                    count += 1
            return count

(2)
class Solution(self,nums,k):
    def findpair(self,nums,k):
        if k < 0:
            return 0
        elif k == 0:
            #collections.Counter(nums).values()为nums中每个元素的个数，是字典的值的形式
            #下面式子统计nums中元素个数大于一的元素总数
            return sum(x > 1 for x in collections.Counter(nums).values())
        else:
            #n+k在nums中的个数
            return len(set(nums)&set(n+k for n in nums))