剑指 Offer 59 - II. 队列的最大值

[Tcdian]

剑指 Offer 59 - II. 队列的最大值

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

Example 1

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

Example 2

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

Note

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

[Tcdian]

Solution

• JavaScript Solution

var MaxQueue = function() {
    this.queue = [];
    this.max_queue = [];
};

/**
 * @return {number}
 */
MaxQueue.prototype.max_value = function() {
    if (this.queue.length > 0) {
        return this.max_queue[0];
    }
    return -1;
};

/** 
 * @param {number} value
 * @return {void}
 */
MaxQueue.prototype.push_back = function(value) {
    this.queue.push(value);
    if (this.max_queue.length > 0) {
        for (let i = this.max_queue.length - 1; i >= 0; i--) {
            if (this.max_queue[i] < value) {
                this.max_queue.pop();
            }
        }
    }
    this.max_queue.push(value);
};

/**
 * @return {number}
 */
MaxQueue.prototype.pop_front = function() {
    if (this.queue.length > 0) {
        const value = this.queue.shift();
        if (this.max_queue[0] === value) {
            this.max_queue.shift();
        }
        return value;
    }
    return -1;
};

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