Given a non-empty array of integers, every element appears twice except for one. Find that single one.
找出列表中只出现一次的数。
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
class Solution:
def singleNumber(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
"""
方法一：
return sum(set(nums))*2 - sum(nums)
"""
ans = 0
for x in nums:
ans = ans ^ x
return ans
def singleNumber(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
"""
方法一：
return sum(set(nums))*2 - sum(nums)
"""
ans = 0
for x in nums:
ans = ans ^ x
return ans