【Single Number】cpp

题目：

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?

代码：

class Solution {
public:
    int singleNumber(vector<int>& nums) {
            int result = 0;
            for (size_t i = 0; i < nums.size(); ++i) result ^= nums[i];
            return result;
    }
};

Tips:

核心思想是用异或运算。

1. 异或运算满足结合律、交换律；结合这两条定律。

2. 出现偶数次的数最终都会对冲掉，成为0；最后剩下那个只出现奇数次的数

3. 任何数与0异或都是其本身

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第二次过这道题，一次AC，考查的是位运算的技巧。

class Solution {
public:
    int singleNumber(vector<int>& nums) {
            int single = 0;
            for ( int i=0; i<nums.size(); ++i ) single  = single ^ nums[i];
            return single;
    }
};