[LeetCode] Implement Queue using Stacks

最新推荐文章于 2021-11-08 15:11:52 发布

转载最新推荐文章于 2021-11-08 15:11:52 发布 · 44 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/maydow/p/4641012.html

本文详细介绍了如何利用两个栈来实现队列的基本操作，包括入队、出队、获取队首元素和判断队列是否为空。通过转换操作减少不必要的元素移动，优化了队列的实现效率。

Implement the following operations of a queue using stacks.

push(x) -- Push element x to the back of queue.
pop() -- Removes the element from in front of queue.
peek() -- Get the front element.
empty() -- Return whether the queue is empty.

Notes:

You must use only standard operations of a stack -- which means only push to top, peek/pop from top, size, and is empty operations are valid.
Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack.
You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).

思路：用两个stack去实现

class MyQueue {
    Stack<Integer> stack1=new Stack<Integer>();  //主
    Stack<Integer> stack2=new Stack<Integer>();   //辅
    // Push element x to the back of queue.
    public void push(int x) {
        stack1.push(x);
    }

    // Removes the element from in front of queue.
    public void pop() {
        while(!stack1.isEmpty())
        {
            stack2.push(stack1.pop());
        }
        if(!stack2.isEmpty())
            stack2.pop();
        while(!stack2.isEmpty())
        {
            stack1.push(stack2.pop());
        }
    }

    // Get the front element.
    public int peek() {
        int ret=0;
        while(!stack1.isEmpty())
        {
            stack2.push(stack1.pop());
        }
        if(!stack2.isEmpty())
        {
            ret=stack2.peek();
        }
        while(!stack2.isEmpty())
        {
            stack1.push(stack2.pop());
        }
        return ret;
    }

    // Return whether the queue is empty.
    public boolean empty() {
        return stack1.isEmpty();
    }
}

这里每次pop和peek操作都会将stack1转到stack2中，比较麻烦。再考虑下可以不用发给转来转去，stack1可以作为队尾，stack2可以作为对首的

class MyQueue {
    Stack<Integer> stack1=new Stack<Integer>();  //主
    Stack<Integer> stack2=new Stack<Integer>();   //辅
    // Push element x to the back of queue.
    public void push(int x) {
        stack1.push(x);
    }

    // Removes the element from in front of queue.
    public void pop() {
        if(!stack2.isEmpty()) stack2.pop();
        else {
            while(!stack1.isEmpty())
            {
                stack2.push(stack1.pop());
            }
            if(!stack2.isEmpty()) stack2.pop();
        }
    }

    // Get the front element.
    public int peek() {
        int ret=0;
        if(!stack2.isEmpty()) ret=stack2.peek();
        else {
            while(!stack1.isEmpty())
            {
                stack2.push(stack1.pop());
            }
            if(!stack2.isEmpty()) ret=stack2.peek();
        }
        return ret;
    }

    // Return whether the queue is empty.
    public boolean empty() {
        return stack1.isEmpty() && stack2.isEmpty();
    }
}