232.Implement Queue using Stacks
===

###### tags: `Easy`,`Stack`,`Queue`

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

### 題目描述

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.

### 範例

**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 `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.

### 解答
#### Python

class MyQueue:

    def __init__(self):
        self.input_stk = []
        self.output_stk = []

    def push(self, x: int) -> None:
        self.input_stk.append(x)

    def pop(self) -> int:
        self.peek()
        return self.output_stk.pop()

    def peek(self) -> int:
        if not self.output_stk:
            while self.input_stk:
                front = self.input_stk.pop()
                self.output_stk.append(front)

        peek = self.output_stk[-1]
        return peek

    def empty(self) -> bool:
        return not self.input_stk and not self.output_stk

>[name=Kobe][time=Fri, Dec 16, 2022]

#### C#

public class MyQueue {
    Stack<int> target = new Stack<int>();
    Stack<int> temp = new Stack<int>();

    public void Push(int x) {
        while(target.Any()){
            temp.Push(target.Pop());
        }
        target.Push(x);
        while(temp.Any()){
            target.Push(temp.Pop());
        }
    }
    
    public int Pop() {
        return target.Pop();
    }
    
    public int Peek() {
        return target.Peek();
    }
    
    public bool Empty() {
        return !target.Any();
    }
}

>[name=Jim][time=Fri, Dec 16, 2022]

#### Javascript

class MyQueue {
  constructor() {
    this.stack = [];
  }

  push(x) {
    this.stack.push(x);
  }

  // use two stacks to implement queue
  pop() {
    const temp = [];
    while (this.stack.length > 1) {
      temp.push(this.stack.pop());
    }

    const result = this.stack.pop();

    while (temp.length > 0) {
      this.stack.push(temp.pop());
    }

    return result;
  }

  peek() {
    return this.stack[0];
  }

  empty() {
    return this.stack.length === 0;
  }
}

>[name=Marsgoat][time= Dec 16, 2022]

#### Java

class MyQueue {

    Stack<Integer> input = new Stack<Integer>();
    Stack<Integer> output = new Stack<Integer>();
    
    public void push(int x) {
        input.push(x);
    }
    
    public int pop() {
        peek();
        return output.pop();
    }
    
    public int peek() {
        if(output.empty()) {
            while(!input.empty()) {
                output.push(input.pop());
            }
        }
        return output.peek();
    }
    
    public boolean empty() {
        return input.empty() && output.empty();
    }
}

>[name=Kobe][time=Fri, Dec 16, 2022]

### Reference

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