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:
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.
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).
Input: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).- 数组里有没有那么几对数，他们之间相差k。
class Solution(object):
def findPairs(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
if k>0:
return len(set(nums) & set(n+k for n in nums))
elif k==0:
return sum(v>1 for v in collections.Counter(nums).values())
else:
return 0
- set()给我带来喜悦，因为他可以去重（chong）。
- 此外，set()还有一些小运算：
>>> a = set('shengfaji')
>>> b = set('kuaileshui')
>>> a
set(['a', 'e', 'g', 'f', 'i', 'h', 'j', 'n', 's'])
>>> a-b # 只在a中存在的元素（差集）
set(['n', 'j', 'g', 'f'])
>>> a|b # a与b的并集
set(['a', 'e', 'g', 'f', 'i', 'h', 'k', 'j', 'l', 'n', 's', 'u'])
>>> a&b # a与b的交集
set(['a', 'i', 's', 'e', 'h'])
>>> a^b # a与b的对称差（（a|b）-（a&b））
set(['g', 'f', 'k', 'j', 'l', 'n', 'u'])
>>>- Counter给我带来喜悦，因为她计数真的很方便.
- 代码
return sum(v>1 for v in collections.Counter(nums).values())作用同下：
flag=0
for v in collections.Counter(nums).values():
if v>1:
flag+=1
return flag- 一般情况下用集合nums&集合nums+k即可解决问题。
- 但是在k=0这种自恋情况下，就只能数数了。
- 综合起来，set和Counter这两份喜悦相互重叠，就能得到了梦幻一般的幸福时光。