# Implement Queue by Two Stacks

## Problem

As the title described, you should only use two stacks to implement a queue's actions.

The queue should support push(element), pop() and top() where pop is pop the first(a.k.a front) element in the queue.

Both pop and top methods should return the value of first element.

## Example

push(1)
pop()     // return 1
push(2)
push(3)
top()     // return 2
pop()     // return 2

## Challenge

implement it by two stacks, do not use any other data structure and push, pop and top should be O(1) by AVERAGE.

## Solution

Queue is first-in-first-out, while stack is first-in-last-out. Since we have two stacks, we can "secretly" reverse the order internally:

• push: add the new element to stack2
• pop/top: if stack1 is not empty, just pop/top from stack1; otherwise keep popping from stack2 and pushing it to stack1 until stack2 is empty. So when calling pop/top from stack1, the elements are in the same order as they are pushed-in

## Code - Java

public class Queue {
private Stack<Integer> stack1;
private Stack<Integer> stack2;

public Queue() {
stack1 = new Stack<>();
stack2 = new Stack<>();
}

public void push(int element) {
stack2.push(element);
}

public int pop() {
if (stack1.isEmpty()) {
move();
}
return stack1.pop();
}

public int top() {
if (stack1.isEmpty()) {
move();
}
return stack1.peek();
}

private void move() {
while (!stack2.isEmpty()) {
stack1.push(stack2.pop());
}
}
}