Leetcode #136 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?

Difficulty:Easy

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int len = nums.size();
        if(len==1)
           return nums[0];
        sort(nums.begin(),nums.end());
        for(int i = 0;i<len;i=i+2)
        {
            if(i==len-1||nums[i]!=nums[i+1])
              return nums[i];
        }
    }
};

Python中,^是位的异或运算，能很巧妙的把那个落单的数找出来

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