136. Single Number Difficulty: Medium

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?

算法分析：a⊕b⊕a=b，每个都去异或，那么最后得到的数就是那个孤立的数。

C语言版

int singleNumber(int* nums, int numsSize) {    int i, n = nums[0];    for(i = 1; i < numsSize; i++)        n ^= nums[i];    return n;}

Python版

class Solution(object):    def singleNumber(self, nums):        """        :type nums: List[int]        :rtype: int        """        n = nums[0]        for t in nums[1:]:            n ^= t        return n