Implement Stack using Queues

描述

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).

分析

可以用两个队列，q和tmp，q存放元素，tmp用来作中转。

• push(x)，先将x push 到tmp，然后把q中的元素全部弹出来，存入tmp，最后切换q和tmp
• pop()，直接将q的队首元素弹出来即可

该算法push的算法复杂度是O(n), pop的算法复杂度O(1)。

另个一个方法是，让pop是O(n), push是O(1)，思路很类似，就不赘述了。

代码

// Implement Stack using Queues
class MyStack {
    // Push element x onto stack.
    // Time Complexity O(n)
    public void push(int x) {
        tmp.offer(x);
        while (!q.isEmpty()) {
            final int e = q.poll();
            tmp.offer(e);
        }
        // swap q and tmp
        Queue temp = tmp;
        tmp = q;
        q = temp;
    }

    // Removes the element on top of the stack.
    // Time Complexity O(1)
    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();
    }

    private Queue q = new LinkedList<>();
    private Queue tmp = new LinkedList<>();
}

相关题目

• Implement Queue using Stacks