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：$a \oplus b \oplus a = b$，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