lc面试准备:Implement Stack using Queues

1 题目

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).

接口: 实现4个方法

2 思路

用2个queue来实现，java推荐用LinkedList作为Queue.

Version A: The stack should be efficient when pushing an item.

Version B: The stack should be efficient when popping an item.

Version A: push O(1); pop O(n)

push:

• enqueue in queue1

pop:

• while size of queue1 is bigger than 1, pipe dequeued items from queue1 into queue2
• dequeue and return the last item of queue1, then switch the names of queue1 and queue2

Version B: push O(n); pop O(1)

push:

• enqueue in queue2
• enqueue all items of queue1 in queue2, then switch the names of queue1 and queue2

pop:

• deqeue from queue1

3 代码

Vesion A

class MyStackVersionA {
	Queue<Integer> q1 = new LinkedList<Integer>();
	Queue<Integer> q2 = new LinkedList<Integer>();

	// Push element x onto stack.
	public void push(int x) {
		q1.add(x);
	}

	// Removes the element on top of the stack.
	public void pop() {
		top();
		q1.poll();
		Queue<Integer> tmp = q1;
		q1 = q2;
		q2 = tmp;
	}

	// Get the top element.
	public int top() {
		int size = q1.size();
		if (size > 1) {
			int count = size - 1;
			for (int i = 0; i < count; i++) {
				q2.add(q1.poll());
			}
		}
		return q1.peek();
	}

	// Return whether the stack is empty.
	public boolean empty() {
		return q1.isEmpty() && q2.isEmpty();
	}
}

Vesion B

class MyStackVersionB {
	Queue<Integer> q1 = new LinkedList<Integer>();
	Queue<Integer> q2 = new LinkedList<Integer>();

	// Push element x onto stack.
	public void push(int x) {
		q2.add(x);
		int size = q1.size();
		for(int i = 0; i < size; i++) {
			q2.add(q1.poll());
		}
		Queue<Integer> tmp = q1;
		q1 = q2;
		q2 = tmp;
	}

	// Removes the element on top of the stack.
	public void pop() {
		q1.poll();
	}

	// Get the top element.
	public int top() {
		return q1.peek();
	}

	// Return whether the stack is empty.
	public boolean empty() {
		return q1.isEmpty() && q2.isEmpty();
	}
}

4 总结

• 用两个queue实现stack, pop 和 push的效率的选择。
• 由于题目假设pop和peek都不会在队列为空的时候执行，避免了Null Pointer Exception.
• stack 和 queue的相互实现，很好的考察基本功。

5 参考

• leetcode
• Implement Stack using Two Queues