225. Implement Stack using Queues**

Implement the following operations of a stack using queues.

• push(x) -- Push element x onto stack.
• pop() -- Removes the element on top of the stack.
• top() -- Get the top element.
• empty() -- Return whether the stack is empty.

Notes:

• You must use only standard operations of a queue -- which means only push to back, peek/pop from front, size, and is empty operations are valid.
• Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.

• You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

One queue:

class MyStack {


    //one Queue solution
    private Queue<Integer> q = new LinkedList<Integer>();
    
    // Push element x onto stack.
    public void push(int x) {
        q.add(x);
        for(int i = 1; i < q.size(); i ++) { //rotate the queue to make the tail be the head
            q.add(q.poll());
        }
    }
    
    // Removes the element on top of the stack.
    public void pop() {
        q.poll();
    }
    
    // Get the top element.
    public int top() {
        return q.peek();        
    }
    
    // Return whether the stack is empty.
    public boolean empty() {
        return q.isEmpty();
    }
}

Two queues

class MyStack {
//using two queue. The push is inefficient.
private Queue<Integer> q1 = new LinkedList<Integer>();
private Queue<Integer> q2 = new LinkedList<Integer>();
public void push(int x) {
    if(q1.isEmpty()) {
        q1.add(x);
        for(int i = 0; i < q2.size(); i ++)
            q1.add(q2.poll());
    }else {
        q2.add(x);
        for(int i = 0; i < q1.size(); i++)
            q2.add(q1.poll());
    }
}

public void pop() {
    if(!q1.isEmpty()) 
        q1.poll();
    else
        q2.poll();
}
public int top() {
    return q1.isEmpty() ? q2.peek() : q1.peek();
}
public boolean empty() {
    return q1.isEmpty() && q2.isEmpty();
}
}

class MyStack{
//using two queue. The pop and top are inefficient.
private Queue<Integer> q1 = new LinkedList<Integer>();
private Queue<Integer> q2 = new LinkedList<Integer>();
public void push(int x) {
    if(!q1.isEmpty()) 
        q1.add(x);
    else
        q2.add(x);
}
public void pop() {
    if(q1.isEmpty()) {
        int size = q2.size();
        for(int i = 1; i < size; i ++) {
            q1.add(q2.poll());
        }
        q2.poll();
    }else {
        int size = q1.size();
        for(int i = 1; i < size; i ++) {
            q2.add(q1.poll());
        }
        q1.poll();
    }
}
public int top() {
    int res;
    if(q1.isEmpty()) {
        int size = q2.size();
        for(int i = 1; i < size; i ++) {
            q1.add(q2.poll());
        }
        res = q2.poll();
        q1.add(res);
    }else {
        int size = q1.size();
        for(int i = 1; i < size; i ++) {
            q2.add(q1.poll());
        }
        res = q1.poll();
        q2.add(res);
    }
    return res;
}
public boolean empty() {
    return q1.isEmpty() && q2.isEmpty();
}
}

总结：其实感觉两个queue没什么必要，还是一个queue比较简洁