题目：

方法一：

使用两个栈，进行push操作时，判断outstack是否空，如果非空，就把Outstack pop到Instack，然后再往instack里面push元素

class MyQueue {
    Stack<Integer> Instack;
    Stack<Integer> Outstack;

    /** Initialize your data structure here. */
    public MyQueue() {
        Instack=new Stack<Integer>();
        Outstack=new Stack<Integer>();
    }
    
    /** Push element x to the back of queue. */
    public void push(int x) {
        while(!Outstack.empty()){
            Instack.push(Outstack.pop());
        }
        Instack.push(x);
        while(!Instack.empty()){
            Outstack.push(Instack.pop());
        }
    }
    
    /** Removes the element from in front of queue and returns that element. */
    public int pop() {
        return Outstack.pop();
    }
    
    /** Get the front element. */
    public int peek() {
        return Outstack.peek();
    }
    
    /** Returns whether the queue is empty. */
    public boolean empty() {
        if(Outstack.empty()){
            return true;
        }else{
            return false;
        }
    }
}

/**
 * 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();
 */

这种方法push时间复杂度为O（n），空间复杂度O（n）
其他方法O（1）

方法二：

再Instack里面，push方法：只要来一个元素，我就给你入栈，
但是pop方法：只有当Outstack是空，我才把Instack里面元素拿出来，不为空就不用管。（因为peek和pop都只是返回头元素，头元素不会因此改变，除非头被干掉）
push时间复杂度：O（1）空间O（n）
pop时间复杂度O（n）空间O（1）
因为pop方法会经历多次Instack倒入Outstack，平均之后还是O（n）

class MyQueue {
    Stack<Integer> Instack;
    Stack<Integer> Outstack;
    Integer y;

    /** Initialize your data structure here. */
    public MyQueue() {
        Instack=new Stack<Integer>();
        Outstack=new Stack<Integer>();
    }
    
    /** Push element x to the back of queue. */
    public void push(int x) {
        if(Instack.empty()){
            y=x;
        }
        Instack.push(x);
    }
    
    /** Removes the element from in front of queue and returns that element. */
    public int pop() {
        if(Outstack.empty()){
            while(!Instack.empty()){
                Outstack.push(Instack.pop());
            }
        }
        return Outstack.pop();
    }
    
    /** Get the front element. */
    public int peek() {
        if(!Outstack.empty()){
            return Outstack.peek();
        }
        return y;
    }
    
    /** Returns whether the queue is empty. */
    public boolean empty() {
        if(Outstack.empty()&&Instack.empty()){
            return true;
        }else{
            return false;
        }
    }
}

/**
 * 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();
 */