# -*- coding: utf-8 -*-

"""

Given a non-empty array of integers, every element appears twice except for one. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Example 1:

Input: [2,2,1]

Output: 1

Example 2:

Input: [4,1,2,1,2]

Output: 4

"""

from collections import Counter

class Solution:

def singleNumber(self, nums):

"""

:type nums: List[int]

:rtype: int

>>> nums = [2,2,1]

>>> s = Solution()

>>> s.singleNumber(nums)

1

>>> nums = [4,1,2,1,2]

>>> s.singleNumber(nums)

4

"""

counter = Counter(nums)

for k, v in counter.items():

if v == 1:

return k

# 这种是不聪明的

# 聪明的是下面这种

# 取原数据集的集合的值乘以2，刚好会比原数据集多出不重复整数的值

# return sum(set(nums)) * 2 - sum(nums)

if __name__ == '__main__':

import doctest

doctest.testmod(verbose=True)