232. Implement Queue using Stacks

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).

一刷
题解：这题跟225的区别是，
用两个stack, input, output
如果仅仅add, 则push到input中去。
如果pop/peek, 如果output为空，则将input中的num全部pop, push到output, 否则只从output中取，不动input

原因是，如果push(1), push(2), peek(), push(3), peek(), 这样就可以保证顺序了。

public class MyStack {
    private Queue q1;
    private Queue q2;

    /** Initialize your data structure here. */
    public MyStack() {
        q1 = new LinkedList<>();
        q2 = new LinkedList<>();
    }
    
    /** Push element x onto stack. */
    public void push(int x) {
        if(q1.isEmpty()){
            q1.add(x);
            for(int i=0; i