[leetcode] 532. K-diff Pairs in an Array @ python

原题

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].

解法

分治法, edge case是当k< 0时, 结果是0. 当k=0时, 计算重复数字的个数, k > 0时, 计算k-diff pairs的个数, 后两者的代码可以合并在一起, 缩短运行时间.
Time: O(n), n是nums中不同数字的个数
Space: O(1)

代码

class Solution(object):
    def findPairs(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        # edge case
        if k < 0: return 0
        ans = 0
        count = collections.Counter(nums)
        for n in count:
            if (k == 0 and count[n] > 1) or (k > 0 and n+k in count):
                ans += 1
                
        return ans