# LeetCode Notes: Implement Queue using Stacks

## Question

Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (`push`

, `peek`

, `pop`

, and `empty`

).

Implement the `MyQueue`

class:

`void push(int x)`

Pushes element x to the back of the queue.`int pop()`

Removes the element from the front of the queue and returns it.`int peek()`

Returns the element at the front of the queue.`boolean empty()`

Returns`true`

if the queue is empty,`false`

otherwise.

**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, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack's standard operations.

**Example 1:**

Input["MyQueue", "push", "push", "peek", "pop", "empty"]

[[], [1], [2], [], [], []]

Output[null, null, null, 1, 1, false]

ExplanationMyQueue myQueue = new MyQueue();

myQueue.push(1); // queue is: [1]

myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)

myQueue.peek(); // return 1

myQueue.pop(); // return 1, queue is [2]

myQueue.empty(); // return false

**Constraints:**

`1 <= x <= 9`

- At most
`100`

calls will be made to`push`

,`pop`

,`peek`

, and`empty`

. - All the calls to
`pop`

and`peek`

are valid.

**Follow-up:** Can you implement the queue such that each operation is **amortized** `O(1)`

time complexity? In other words, performing `n`

operations will take overall `O(n)`

time even if one of those operations may take longer.

For questions of the same type, please refer to

## Solution

**Analysis:**

Take advantage of the characteristics of the js array, where the top of the stack is at the head of the array.

**Code:**

```
/**
* Initialize your data structure here.
*/
var MyQueue = function() {
this.queue = []
};
/**
* Push element x to the back of queue.
* @param {number} x
* @return {void}
*/
MyQueue.prototype.push = function(x) {
this.queue.push(x)
};
/**
* Removes the element from in front of queue and returns that element.
* @return {number}
*/
MyQueue.prototype.pop = function() {
return this.queue.shift()
};
/**
* Get the front element.
* @return {number}
*/
MyQueue.prototype.peek = function() {
return this.queue[0]
};
/**
* Returns whether the queue is empty.
* @return {boolean}
*/
MyQueue.prototype.empty = function() {
return this.queue.length === 0
};
/**
* Your MyQueue object will be instantiated and called as such:
* var obj = new MyQueue()
* obj.push(x)
* var param_2 = obj.pop()
* var param_3 = obj.peek()
* var param_4 = obj.empty()
*/
```

## Comments