[LeetCode][JavaScript]Implement Queue using Stacks

Implement Queue using Stacks 

Implement the following operations of a queue using stacks.

• push(x) -- Push element x to the back of queue.
• pop() -- Removes the element from in front of queue.
• peek() -- Get the front element.
• empty() -- Return whether the queue is empty.

Notes:

• You must use only standard operations of a stack -- which means only push to top, peek/pop from top, size, and is empty operations are valid.
• Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
• You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).

 https://leetcode.com/problems/implement-queue-using-stacks/

 

 

 

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2个栈实现队列。

栈1为主，栈2为辅。

push操作，先把栈2所有的元素都倒入栈1，然后对栈1压桟。

pop和peek操作，如果栈2为空，就把栈1所有元素都倒入栈2，然后对栈2出桟。

/**
 * @constructor
 */
var Queue = function() {
    this.stack1 = [];
    this.stack2 = [];
};

/**
 * @param {number} x
 * @returns {void}
 */
Queue.prototype.push = function(x) {
    var len = this.stack2.length;
    while(len--){
        this.stack1.push(this.stack2.pop());
    }
    this.stack1.push(x);
};

/**
 * @returns {void}
 */
Queue.prototype.pop = function() {
    if(this.stack2.length === 0){
        var len = this.stack1.length;
        while(len--){
            this.stack2.push(this.stack1.pop());
        }
    }
    return this.stack2.pop();
};

/**
 * @returns {number}
 */
Queue.prototype.peek = function() {
    if(this.stack2.length === 0){
        var len = this.stack1.length;
        while(len--){
            this.stack2.push(this.stack1.pop());
        }
    }
    return this.stack2[this.stack2.length - 1];
};

/**
 * @returns {boolean}
 */
Queue.prototype.empty = function() {
    if(this.stack1.length === 0 && this.stack2.length === 0){
        return true;
    }
    return false;
};