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Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.37

232\. Implement Queue using Stacks

Easy

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.