博文中会简要介绍Leetcode P0136题目分析及解题思路。

“Single Number”这道题比较简单，可以用HashTable来完成，也可以通过位运算的技巧，由于两个相同的数异或后的结果是0，所以把所有数都异或以后，唯一的那个数就会剩下。

Given a non-empty 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?

以下是Java的题解代码实现。

class Solution {
    public int singleNumber(int[] nums) {
        int single = nums[0];
        for (int idx=1; idx<nums.length; ++idx) 
            single ^= nums[idx];
        
        return single;
    }
}

以下是C++的题解代码实现。

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int single = nums[0];
        for (int idx=1; idx<nums.size(); ++idx)
            single ^= nums[idx];
        
        return single;
    }
};