Programmer interview book, cat and dog shelter, programmer interview book

Programmer interview book, cat and dog shelter, programmer interview book

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Question link: http://www.nowcoder.com/practice/6235a76b1e404f748f7c820583125c50? Rp = 1 & ru =/ta/cracking-the-coding-interview & qru =/ta/cracking-the-coding-interview/question-ranking

Description
There is a home animal shelter that only supports cats and dogs, but there are special adoption rules. There are two adoption methods for the recipient. The first is to directly adopt all the animals to enter the shelter, the second type is the type of the adopted animal (cat or dog), and the earliest adoption of the animal to enter the shelter.
Given an operation sequence int [] [2] ope (vector <int> in C ++) represents all events. If the first element is 1, it indicates that an animal enters the shelter. The second element indicates the number of the animal. An integer indicates the dog. A negative number indicates the cat. If the first element is 2, it indicates that an animal is adopted. If the second element is 0, the first adoption mode is adopted. If it is 1, the dog is adopted. If it is-1, the cat is adopted. Return the adoption sequence in order. If there is an illegal operation, that is, there are no animals that can meet the adoption requirements, the adoption operation will be ignored.
Test example:
[[], [1,-1], [], [2,-1]
Return Value: [1,-1]

Ideas
We use two queues to complete all operations. For the first operation, we only need to return the first element in the queue.
For the second operation, we store all the data that does not meet the requirements into the auxiliary queue, and find the first element that meets the conditions, then delete it, then, all the elements in the auxiliary queue are restored to the queue.

class CatDogAsylum{    public:        vector<int> asylum(vector<vector<int> > ope)        {            // write code here            vector<int> ans;            int len = ope.size();            if(len==0)                return ans;            queue<int> Q1;            queue<int> Q2;            for(int i = 0; i<len; i++)            {                if(ope[i][0]==1)                {                    Q1.push(ope[i][1]);                }                else                {                    if(ope[i][1]==0)                    {                        if(!Q1.empty())                        {                            ans.push_back(Q1.front());                            Q1.pop();                        }                    }                    else                    {                        while(!Q1.empty())                        {                            if(ope[i][1]<0 && Q1.front()<0)                            {                                ans.push_back(Q1.front());                                Q1.pop();                                break;                            }                            else if(ope[i][1]>0 && Q1.front()>0)                            {                                ans.push_back(Q1.front());                                Q1.pop();                                break;                            }                            else                            {                                Q2.push(Q1.front());                                Q1.pop();                            }                        }                        while(!Q1.empty())                        {                            Q2.push(Q1.front());                            Q1.pop();                        }                        while(!Q2.empty())                        {                            Q1.push(Q2.front());                            Q2.pop();                        }                    }                }            }            return ans;        }};

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