# 0232. Implement Queue using Stacks

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

## Description

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]
**Explanation**
MyQueue 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**](https://en.wikipedia.org/wiki/Amortized_analysis) `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.

## ac

```java
class MyQueue {
    Stack<Integer> stack, stack2;
    /** Initialize your data structure here. */
    public MyQueue() {
        stack = new Stack<>();
        stack2 = new Stack<>();
    }

    /** Push element x to the back of queue. */
    public void push(int x) {
        stack.push(x);
    }

    /** Removes the element from in front of queue and returns that element. */
    public int pop() {
        if (stack2.isEmpty()) {
            while (!stack.isEmpty())
                stack2.push(stack.pop());
        }
        return stack2.pop();
    }

    /** Get the front element. */
    public int peek() {
        if (stack2.isEmpty()) {
            while (!stack.isEmpty())
                stack2.push(stack.pop());
        }
        return stack2.peek();
    }

    /** Returns whether the queue is empty. */
    public boolean empty() {
        return stack.isEmpty() && stack2.isEmpty();
    }
}

/**
 * Your MyQueue object will be instantiated and called as such:
 * MyQueue obj = new MyQueue();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.peek();
 * boolean param_4 = obj.empty();
 */

/*
just 2 stack, when pop()/peek(), pour all elements from stack1 to stack2
*/
```
