剑指 Offer 59

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

示例 1：

输入: 
["MaxQueue","push_back","push_back","max_value","pop_front","max_value"]-。0
[[],[1],[2],[],[],[]]
输出: [null,null,null,2,1,2]

示例 2：

输入: 
["MaxQueue","pop_front","max_value"]
[[],[],[]]
输出: [null,-1,-1]

限制：

• 1 <= push_back,pop_front,max_value的总操作数 <= 10000
• 1 <= value <= 10^5

class MaxQueue {
    queue<int>q;
    deque<int>d;
public:
    MaxQueue() {

    }
    
    int max_value() {
        if(d.empty())return -1;
        return d.front();
    }
    
    void push_back(int value) {
        while (!d.empty() && d.back() < value) {//保持front()最大
            d.pop_back();
        }
        d.push_back(value);
        q.push(value);
    }

    int pop_front() {
        if (q.empty())return -1;
        int res = q.front();
        if (res == d.front()) {//注意一起pop
            d.pop_front();
        }
        q.pop();
        return res;
    }
};

/**
 * Your MaxQueue object will be instantiated and called as such:
 * MaxQueue* obj = new MaxQueue();
 * int param_1 = obj->max_value();
 * obj->push_back(value);
 * int param_3 = obj->pop_front();
 */