Description

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).

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and all test cases.

Solution

Two queues, push O(n), other functions O(1)

每次push时把所有元素折腾成逆进入顺序的。感觉有点像递归啊：
reverse(a, n)代表把a中n各元素调整成逆序，用递归来做就是返回a[n] append reverse(a, n - 1)。

class MyStack {
    private Queue<Integer> q1;
    private Queue<Integer> 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.offer(x);
            while (!q2.isEmpty()) {
                q1.offer(q2.poll());
            }
        } else {
            q2.offer(x);
            while (!q1.isEmpty()) {
                q2.offer(q1.poll());
            }
        }
    }
    
    /** Removes the element on top of the stack and returns that element. */
    public int pop() {
        return q1.isEmpty() ? q2.poll() : q1.poll();
    }
    
    /** Get the top element. */
    public int top() {
        return q1.isEmpty() ? q2.peek() : q1.peek();
    }
    
    /** Returns whether the stack is empty. */
    public boolean empty() {
        return q1.isEmpty() && q2.isEmpty();
    }
}

/**
 * Your MyStack object will be instantiated and called as such:
 * MyStack obj = new MyStack();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.top();
 * boolean param_4 = obj.empty();
 */

Optimized: One queue, push O(n), other functions O(1)

其实一个queue就够了啦。

class MyStack {
    Queue<Integer> q;
    /** Initialize your data structure here. */
    public MyStack() {
        q = new LinkedList<>();
    }
    
    /** Push element x onto stack. */
    public void push(int x) {
        q.offer(x);
        for (int i = 0; i < q.size() - 1; ++i) {
            q.offer(q.poll());  // rotate prev elements to the tail
        }
    }
    
    /** Removes the element on top of the stack and returns that element. */
    public int pop() {
        return q.poll();
    }
    
    /** Get the top element. */
    public int top() {
        return q.peek();
    }
    
    /** Returns whether the stack is empty. */
    public boolean empty() {
        return q.isEmpty();
    }
}

/**
 * Your MyStack object will be instantiated and called as such:
 * MyStack obj = new MyStack();
 * obj.push(x);
 * int param_2 = obj.pop();
 * int param_3 = obj.top();
 * boolean param_4 = obj.empty();
 */