136. Single Number

                             Given an array of integers, every element appears twice except for one. Find that single one.

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

class Solution { // use XORpublic:    int singleNumber(vector<int>& nums) {int n = nums.size();if (n == 0){return -1;}        int result = nums[0];        for (int i = 1; i < nums.size(); i++)        {            result = result ^ nums[i];        }        return result;    }};

第二种方法很神奇。 1001^1001 is 0000.

9,3,9. ：

step 1: 1001^0011 = 1010
step 2: 1010^1001 = 0011