面试题59 - II. 队列的最大值

Tags: Medium Queue

Links: https://leetcode-cn.com/problems/dui-lie-de-zui-da-zhi-lcof/

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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"]

[[],[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

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

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

    void push_back(int value) {
        q.push(value);
        while (!dq.empty() && dq.back() < value) {
            dq.pop_back();
        }
        dq.push_back(value);
    }

    int pop_front() {
        if (q.empty()) return -1;
        int res = q.front();
        q.pop();
        if (res == dq.front()) dq.pop_front();
        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();
 */