Implement Stack using Queues Leetcode

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).

 

这道题可以用一个queue实现，我却用了两个。。。两个queue的思路就是一个queue里面放顶点，另一个queue里面按照stack顺序存储。

代码写的很长。。。

public class MyStack {
    Queue<Integer> q1;
    Queue<Integer> q2;
    boolean isQ1;

    /** Initialize your data structure here. */
    public MyStack() {
        q1 = new LinkedList<>();
        q2 = new LinkedList<>();
        isQ1 = true;
    }
    
    /** Push element x onto stack. */
    public void push(int x) {
        if (q1.isEmpty()) {
            q1.offer(x);
            isQ1 = true;
        } else if (q2.isEmpty()) {
            q2.offer(x);
            isQ1 = false;
        } else if (isQ1) {
            while (!q2.isEmpty()) {
                q1.offer(q2.poll());
            }
            q2.offer(x);
            isQ1 = false;
        } else {
            while (!q1.isEmpty()) {
                q2.offer(q1.poll());
            }
            q1.offer(x);
            isQ1 = true;
        }
    }
    
    /** Removes the element on top of the stack and returns that element. */
    public int pop() {
        int x = 0;
        if (isQ1) {
            x = q1.poll();
            if (!q2.isEmpty()) {
                q1.offer(q2.poll());
            }
        } else {
            x = q2.poll();
            if (!q1.isEmpty()) {
                q2.offer(q1.poll());
            }
        }
        return x;
    }
    
    /** Get the top element. */
    public int top() {
        if (isQ1) {
            return q1.peek();
        }
        return q2.peek();
    }
    
    /** Returns whether the stack is empty. */
    public boolean empty() {
        if (q1.size() == 0 && q2.size() == 0) {
            return true;
        }
        return false;
    }
}

/**
 * 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();
 */

 

一个queue的思路就是每次enqueue， dequeue。

public 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());
        }
    }
    
    /** 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();
 */

这个算法果然跑得很快。。。只有push的时候是O(n)。因为push的时候就是按照stack的顺序push进去的。