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

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 @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