题目
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
解答:
因为只有一个数字是出现了一次的,其他的都是出现了两次,那么可以用按位异或,因为任意一个数字和其自己按位异或得到的结果都是0.
所以最后的那个数就是结果
class Solution(object):
def singleNumber(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
x = nums[0]
for y in nums[1:]:
x ^= y
return x





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