[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:

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51

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)

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43

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)

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

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