# Implement Queue using Stacks Solutions in Java

Number 232

Difficulty Easy

Acceptance 49.7%

Link LeetCode

## Solutions

### Java solution by haoel/leetcode

// Source : https://leetcode.com/problems/implement-queue-using-stacks/description/// Author : Tianming Cao// Date : 2018-02-02package myQueue;import java.util.Stack;/*** This problem is a sibling of problem 225(https://leetcode.com/problems/implement-stack-using-queues/description/)* The solution is:* 1. stack1 is always in charge of push operation* 2. stack2 is always in charge of peek and pop operation* 3. if we want to do peek or pop operation, but stack2 is empty,* we should first pop all the elements from stack1 and push them into stack2 in turn.* Give a Example:** First, push numbers "1,2,3,4,5" to stack1, then stack1's structure is:** |5|* |4|* |3|* |2|* |1|** Second, if we want to get the front element "1",we should pop all the elements of stack1 and push them into stack2,* after this, stack1 is empty, and stack2's structrue is:** |1|* |2|* |3|* |4|* |5|** So we can get stack2's top element "1" as the front element of queue.** Next, if we want to push "6" to the back of queue, we should push "6" into stack1 as before, so stack1's structure is:** |6|** Finally, if we want to do pop operation twice ,we should remove the top element of stack2 twice, so stack2's structure is:** |3|* |4|* |5|** as expect, the removed element is "1" and "2".*/public class MyQueue {public Stack<Integer> stack1;public Stack<Integer> stack2;public int size;public MyQueue() {stack1 = new Stack<>();stack2 = new Stack<>();size = 0;}public void push(int x) {stack1.push(x);size++;}public int pop() {if (stack2.isEmpty()) {while (!stack1.isEmpty()) {stack2.push(stack1.pop());}}int value = stack2.pop();size--;return value;}public int peek() {if (stack2.isEmpty()) {while (!stack1.isEmpty()) {stack2.push(stack1.pop());}}return stack2.peek();}public boolean empty() {return size == 0 ? true : false;}}