Leetcode-队列得最大值

请定义一个队列并实现函数 max_value 得到队列里的最大值，要求函数max_value、push_back 和 pop_front 的均摊时间复杂度都是O(1)。若队列为空，pop_front 和 max_value 需要返回 -1

O(1) 复杂度，意味着min值需要采取特殊手段：双端队列：

• 当push的值大于队尾元素，则pop掉队列中比这个值大的元素，再把这个元素入队
• 否则，直接入队

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <climits>

using namespace std;

/*
tail = 0,head=0,max_size;
push:
pop:
empty:
full:
*/

class Dequque{
private:
    int *data;
    int max_size;
    int head_loc;
    int tail_loc;
public:
    Dequque();
    void push(const int value);
    void pop_back();
    void pop_front();
    int front();
    int tail();
    bool empty();
    bool full();
};

Dequque::Dequque(){
    max_size = 10000;
    head_loc = 0;
    tail_loc = 0;
    data = new int[max_size];
}

void Dequque::push(const int value){
    if(full())
        std::cerr<<"full, cant push"<<endl;
    data[tail_loc] = value;
    tail_loc = (tail_loc+1) % max_size;
}

void Dequque::pop_back(){
    if(empty())
        std::cerr<<"empty, cant pop back"<<endl;
    tail_loc = (tail_loc-1) % max_size;
}

void Dequque::pop_front(){
    if(empty())
        std::cerr<<"empty, cant pop front"<<endl;
    head_loc = (head_loc+1) % max_size;

}

int Dequque::front(){
    if(!empty())
        return data[head_loc];
    return -1;
}

int Dequque::tail(){
    if(!empty())
        return data[tail_loc-1];
    return -1;
}

bool Dequque::empty(){
    return head_loc==tail_loc;
}

bool Dequque::full(){
    return (tail_loc+1) % max_size == head_loc;
}

class MaxQueue {
private:
    int *data;
    int max_size;
    int head;
    int tail;

    Dequque dequeue;

public:
    MaxQueue() {
        data = new int[10000];
        max_size = 10000;
        head = 0;
        tail = 0;
    }

    int max_value_slow(){
        int max_val = INT_MIN;
        if(head==tail)
            return -1;
        for(int i=head;i<=tail;i++)
            max_val = max(max_val, data[i]);
        return max_val;
    }

    int max_value(){
        if(!dequeue.empty())
            return dequeue.front();
        return -1;
    }

    void push_back(int value) {
        data[tail] = value;
        tail = (tail+1)%max_size;

        if(dequeue.empty() || dequeue.tail()>value)
            dequeue.push(value);
        else{
            while(!dequeue.empty() && dequeue.tail()<value)
                dequeue.pop_back();
            dequeue.push(value);
        }
    }

    int pop_front(){
        if(empty())
            return -1;
        int val = front();
        head = (head+1)%max_size;

        if(val==dequeue.front())
            dequeue.pop_front();

        return val;
    }

    int front(){
        if(!empty())
            return data[head];
        else
            return -1;
    }

    bool empty(){
        return head==tail;
    }
};