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 back, pop from front, size, and is empty operations are valid.

 

Credits:

Special thanks to  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
     
       queue1 = new LinkedList
      
       ();    LinkedList
       
         queue2 = new LinkedList
        
         ();     // 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) == 0class 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：