[LeetCode] Challenge log 136

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136. Single Number
  1. Use hash table to mark element that shows before.
  2. Use math, 2 * sum of all unique number - sum of all.
  3. Use a XOR 0 = a, a XOR a =0.

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:

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Input: [2,2,1]
Output: 1

Example 2:

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Input: [4,1,2,1,2]
Output: 4
Soulution:
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# s1
class Solution:
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        uni = set()
        while len(nums) > 0:
            num = nums.pop()
            if num in uni: 
                uni.remove(num)
            else:
                uni.add(num)
        return uni.pop()
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# s2
class Solution:
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return 2 * sum(set(nums)) - sum(nums)
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# s3
class Solution:
    def singleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        for i in range(1, len(nums)):
            nums[0] = nums[0] ^ nums[i]
        return nums[0]
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