[勇者闯LeetCode] 136. Single Number

Description

Given an 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?

Information

Tags: Hash Table | Bit Manipulation
Difficulty: Easy

Solution

两个相同的数字进行异或操作会得到0，而任一数字与0进行异或则得到原来的数字。由于数组只一个数字只出现一次，其他都出现两次，所以对数组中的数字逐个进行异或操作，最后会得到只出现一次的那个数字。

Python Code

class Solution(object):def singleNumber(self, nums):""":type nums: List[int]:rtype: int"""result = 0for num in nums:result ^= numreturn result

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[勇者闯LeetCode] 136. Single Number

[勇者闯LeetCode] 136. Single Number

Description

Information

Solution

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