# 232 Implement Queue using Stacks

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

## Question

Implement the following operations of a queue using stacks.

* push(x) -- Push element x to the back of queue.
* pop() -- Removes the element from in front of queue.
* peek() -- Get the front element.
* empty() -- Return whether the queue is empty.

**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, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
* You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).

## Implementation

思路：主要是需要两个stack来保证输入的顺序和queue的一致

```java
class MyQueue {
    Stack<Integer> oldStack;
    Stack<Integer> newStack;

    /** Initialize your data structure here. */
    public MyQueue() {
        oldStack = new Stack<>();
        newStack = new Stack<>();
    }

    /** Push element x to the back of queue. */
    public void push(int x) {
        while (!oldStack.isEmpty()) {
            newStack.push(oldStack.pop());
        }
        newStack.push(x);
    }

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

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

    /** Returns whether the queue is empty. */
    public boolean empty() {
        return oldStack.isEmpty() && newStack.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();
 */
```

## Time & Space Complexity

所有操作的时间复杂度都是O(n), 空间复杂度是O(n)


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