Python3解leetcode Single Number
问题描述:
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:
考虑异或运算, 例如5^6,就是101^110,结果是011.即两个完全相同的数字进行异或运算,得到的结果为0,。
根据以上特性,将所有num中的数字依次进行异或运算,则其中重复的项经过运算为0,最终结果即单个的项
又由于尽量不适用其他内存,因而考虑将每一步异或结果放置于num[0]的位置
代码:
class Solution:
def singleNumber(self, nums: List[int]) -> int:
for i in nums[1:]:
nums[0] = i^nums[0]
return nums[0]
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