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面试题59 - II. 队列的最大值

题目描述

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

解法

利用一个辅助队列按单调顺序存储当前队列的最大值。

Python3

class MaxQueue:

    def __init__(self):
        self.p = []
        self.q = []

    def max_value(self) -> int:
        return -1 if not self.q else self.q[0]

    def push_back(self, value: int) -> None:
        while self.q and self.q[-1] < value:
            self.q.pop(-1)
        self.q.append(value)
        self.p.append(value)
        

    def pop_front(self) -> int:
        if not self.p: return -1
        res = self.p.pop(0)
        if res == self.q[0]:
            self.q.pop(0)
        return res

Java

class MaxQueue {

    private Queue<Integer> p = new ArrayDeque<>();
    private Deque<Integer> q = new ArrayDeque<>();

    public MaxQueue() {

    }
    
    public int max_value() {
        return q.isEmpty() ? -1 : q.peekFirst();
    }
    
    public void push_back(int value) {
        while (!q.isEmpty() && q.peekLast() < value) {
            q.pollLast();
        }
        q.addLast(value);
        p.add(value);
    }
    
    public int pop_front() {
        if (p.isEmpty()) {
            return -1;
        }
        int res = p.poll();
        if (res == q.peekFirst()) {
            q.pollFirst();
        }
        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();
 */

JavaScript

var MaxQueue = function() {
    this.queue = []
    this.maxValue = -Infinity
    this.maxIdx = -1
};

/**
 * @return {number}
 */
MaxQueue.prototype.max_value = function() {
    if(!this.queue.length) return -1
    return this.maxValue
};

/** 
 * @param {number} value
 * @return {void}
 */
MaxQueue.prototype.push_back = function(value) {
    this.queue.push(value)
    if(value >= this.maxValue) {
        this.maxIdx = this.queue.length-1
        this.maxValue = value
    }
};

/**
 * @return {number}
 */
MaxQueue.prototype.pop_front = function() {
    if(!this.queue.length) return -1
    let a = this.queue.shift()
    this.maxIdx--
    if(this.maxIdx < 0) {
        let tmp = -Infinity
        let id = -1
        for(let i=0;i<this.queue.length;i++) {
            if(this.queue[i] > tmp) {
                tmp = this.queue[i]
                id = i
            }
        }
        this.maxIdx = id
        this.maxValue = tmp
    }
    return a
};

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