## [232\. Implement Queue using Stacks](https://leetcode.com/problems/implement-queue-using-stacks/)

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.

解法：使用兩個 stacks


class MyQueue {
public:
    MyQueue() {}
    
    void push(int x) {
        // 後進來的會維持在最上層
        // 如果 s1 推了 1, 2, 3，3 會在最上面
        s1.push(x);
    }
    
    int pop() {
        peek();
        int temp = s2.top(); s2.pop();
        return temp;
    }
    
    int peek() {
        // s2 用來儲存 s1 的反轉順序，如果 s2 裡面有值，直接回傳 s2.top();
        if(!s2.empty()) return s2.top();
        // 如果 s1 還有值，將元素丟到 s2 儲存，反轉順序
        while(!s1.empty()) {
            s2.push(s1.top()); s1.pop();
        }
        return s2.top();
    }
    
    bool empty() {
        return s1.empty() && s2.empty();
    }
private:
    stack<int> s1; // push 進去的元素先裝在這，最後一個元素會維持在最上層
    stack<int> s2; // 將 s1 的元素 pop() 到 s2，反轉順序
};

/**
 * 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();
 * bool param_4 = obj->empty();
 */

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(N)$
:::