面试题59 - II

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"]
[[],[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

Solutions

1. mono queue

class MaxQueue {
    deque<int> q;
    deque<int> maxq;

public:
    MaxQueue() {

    }

    int max_value() {
        if (!q.size())
            return -1;
        else
            return maxq.front();        
    }

    void push_back(int value) {
        q.push_back(value);
        while (maxq.size() && maxq.back() < value)
            maxq.pop_back();
        maxq.push_back(value);
    }

    int pop_front() {
        if (!q.size())
            return -1;
        int front = q.front();
        q.pop_front();
        if (front == maxq.front())
            maxq.pop_front();
        return front;
    }
};

/**
 * 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();
 */

Or use treemap to maintain the maximum value.