260.Single Number III

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:

1. The order of the result is not important. So in the above example, [5, 3] is also correct.
2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

思路：如果没有note中2的要求，那么很简单，用循环找出这两个数并装入vector容器中即可，可是超时了。

 

我的代码：

class Solution {
public:
    vector<int> singleNumber(vector<int>& nums) 
    {
        int result[2]={0,0};
        for(int i=0, status=0,count=0;i<nums.size();i++)
        {
            for(int j=0;j<nums.size();j++)
            {
                if ((nums[i]==nums[j])&&(i!=j))
                status=1;
            }
            if (status == 0)
            {
                result[count]=nums[i];
                count++;
            }
        }
        int A=result[0],B=result[1];
        return vector<int>({A,B});
    }
};

 

 

参考代码：

class Solution {
public:
    vector<int> singleNumber(vector<int>& nums) {
        int AxorB = 0;
        for(int i = 0;i<nums.size();i++){
            AxorB^=nums[i];
        }
        //取最后一个二进制位
        int mask = AxorB & (~(AxorB-1));
        int A = 0,B = 0;
        for(int i = 0;i<nums.size();i++){
            if(mask&nums[i])
                A ^= nums[i];
            else
                B ^= nums[i];
        }

        return vector<int>({A,B});
    }
};

接下来这几天要开发山寨一款新浪微博iOS版，所以刷题停几天，正好也要学swift，以后就用swift刷试试看，并且推到github上，接下来blog更新内容就是微博每一天开发的进度。