Problem
Given an array of integers, every element appears twice except for one. Find that single one.
Solution
?可用嵌套循環(huán)暴力求解，時間復(fù)雜度為O(n2)。
?用異或運(yùn)算巧解，時間復(fù)雜度為O(n)。
class Solution {
public:
int singleNumber(vector<int>& nums) {
int result = 0;
for (int i = 0; i < nums.size(); i++){
result ^= nums[i];
}
return result;
}
};
?原理：輸入數(shù)組 [4, 2, 4, 1, 2, 3, 3]
?運(yùn)算等同于 (2 ^ 2) ^ (3 ^ 3) ^ (4 ^ 4) ^ 1 = 0 ^ 0 ^ 0 ^ 1 = 1
