[LeetCode] 225. Implement Stack using Queues 用队列来实现栈

Posted 轻风舞动

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了[LeetCode] 225. Implement Stack using Queues 用队列来实现栈相关的知识,希望对你有一定的参考价值。

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 may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).
  • 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 -- which means only push to backpop from frontsize, and is empty operations are valid.

 

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

用队列Queues的基本操作来实现栈Stack,队列和栈都是重要的数据结构,区别主要是,队列是先进先出,而栈是先进后出。

解法1:  在队列push进来新元素时,就把其它元素pop出来排到新元素的后面,新元素就在前面了,就可以后进先出。就好像大家都在排队,来了个重要客人,要让他第一,其他人从前面按顺序跑到他后面。

解法2: push的时候,其他元素不动只是用一个变量记住这个新元素。当top的时候直接给这个变量的值。当pop时,在调整顺序,把最后一个排到前面,弹出。变量记住目前在最尾部的值。

Java:

class MyStack {
    LinkedList<Integer> queue1 = new LinkedList<Integer>();
    LinkedList<Integer> queue2 = new LinkedList<Integer>();
 
    // Push element x onto stack.
    public void push(int x) {
        if(empty()){
            queue1.offer(x);
        }else{
            if(queue1.size()>0){
                queue2.offer(x);
                int size = queue1.size();
                while(size>0){
                    queue2.offer(queue1.poll());
                    size--;
                }
            }else if(queue2.size()>0){
                queue1.offer(x);
                int size = queue2.size();
                while(size>0){
                    queue1.offer(queue2.poll());
                    size--;
                }
            }
        }
    }
 
    // Removes the element on top of the stack.
    public void pop() {
        if(queue1.size()>0){
            queue1.poll();
        }else if(queue2.size()>0){
            queue2.poll();
        }
    }
 
    // Get the top element.
    public int top() {
       if(queue1.size()>0){
            return queue1.peek();
        }else if(queue2.size()>0){
            return queue2.peek();
        }
        return 0;
    }
 
    // Return whether the stack is empty.
    public boolean empty() {
        return queue1.isEmpty() & queue2.isEmpty();
    }
} 

Python: Time: push: O(n), pop: O(1), top: O(1), Space: O(n)

class Queue:
    def __init__(self):
        self.data = collections.deque()
        
    def push(self, x):
        self.data.append(x)
    
    def peek(self):
        return self.data[0]
    
    def pop(self):
        return self.data.popleft()
    
    def size(self):
        return len(self.data)
    
    def empty(self):
        return len(self.data) == 0


class Stack:
    # initialize your data structure here.
    def __init__(self):
        self.q_ = Queue()

    # @param x, an integer
    # @return nothing
    def push(self, x):
        self.q_.push(x)
        for _ in xrange(self.q_.size() - 1):
            self.q_.push(self.q_.pop())

    # @return nothing
    def pop(self):
        self.q_.pop()

    # @return an integer
    def top(self):
        return self.q_.peek()

    # @return an boolean
    def empty(self):
        return self.q_.empty()

Python:  Time: push: O(1), pop: O(n), top: O(1),Space: O(n)

class Stack:
    # initialize your data structure here.
    def __init__(self):
        self.q_ = Queue()
        self.top_ = None

    # @param x, an integer
    # @return nothing
    def push(self, x):
        self.q_.push(x)
        self.top_ = x

    # @return nothing
    def pop(self):
        for _ in xrange(self.q_.size() - 1):
            self.top_ = self.q_.pop()
            self.q_.push(self.top_)
        self.q_.pop()

    # @return an integer
    def top(self):
        return self.top_

    # @return an boolean
    def empty(self):
        return self.q_.empty()

 

类似题目:

 [LeetCode] 232. Implement Queue using Stacks 用栈来实现队列

All LeetCode Questions List 题目汇总

  

  

以上是关于[LeetCode] 225. Implement Stack using Queues 用队列来实现栈的主要内容,如果未能解决你的问题,请参考以下文章

leetcode?python 225. Implement Stack using Queues

LeetCode 225. Implement Stack using Queues

LeetCode 225 Implement Stack using Queues

[leetcode] 225. Implement Stack using Queues

LeetCode OJ 225Implement Stack using Queues

Leetcode 225 Implement Stack using Queues STL