难度: Easy
原题连接
内容描述
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
思路 1 - 时间复杂度: O(N)- 空间复杂度: O(N)******
位运算,终于要take it了
非常常见的一道算法题,将所有数字进行异或操作即可。对于异或操作明确以下三点:
- 一个整数与自己异或的结果是0
- 一个整数与0异或的结果是自己
- 异或操作满足交换律,即a^b=b^a
Python的位操作: https://wiki.python.org/moin/BitwiseOperators
class Solution(object):
def singleNumber(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
res = nums[0]
for i in nums[1:]:
res ^= i
return res