【LeetCode】Single Number

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?

 

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Solution：

解法不少，贴一种：

1 class Solution:

2     # @param {integer[]} nums

3     # @return {integer}

4     def singleNumber(self, nums):

5         ans = nums[0];

6         for i in range(1, len(nums)):

7             ans ^= nums[i]

8         return ans

 

其依据在于：

1、异或运算满足交换律；

2、a ^ a = 0;

3、b ^ 0 = b。

 

这题的关键就在于线性时间内把相同的一对 “消掉”，留下那个 “落单” 的。

异或运算给了这样的快捷的可能。