LeetCode 225. Implement Stack using Queues

题目

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 must use only standard operations of a queue – which means only push to back, peek/pop from front, size, and is empty operations are valid.

  • 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.
  • You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

分析

用队列(列表)实现一个栈

代码

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class MyStack(object):
def __init__(self):
"""
Initialize your data structure here.
"""
self._queue = collections.deque()
def push(self, x):
"""
Push element x onto stack.
:type x: int
:rtype: void
"""
q = self._queue
q.append(x)
for _ in range(len(q) - 1):
q.append(q.popleft())
def pop(self):
"""
Removes the element on top of the stack and returns that element.
:rtype: int
"""
return self._queue.popleft()
def top(self):
"""
Get the top element.
:rtype: int
"""
return self._queue[0]
def empty(self):
"""
Returns whether the stack is empty.
:rtype: bool
"""
return not len(self._queue)
# class Stack:
# # initialize your data structure here.
# def __init__(self):
# self.stack = collections.deque([])
# # @param x, an integer
# # @return nothing
# def push(self, x):
# self.stack.append(x)
# # @return nothing
# def pop(self):
# for i in range(len(self.stack) - 1):
# self.stack.append(self.stack.popleft())
# self.stack.popleft()
# # @return an integer
# def top(self):
# return self.stack[-1]
# # @return an boolean
# def empty(self):
# return len(self.stack) == 0
# Your MyStack object will be instantiated and called as such:
# obj = MyStack()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.top()
# param_4 = obj.empty()

点评

两种方法,看是需要牺牲 pop 还是 push 的时间复杂度
可不可以牺牲空间,让 pop 和 push 都是 常数时间呢?好像没有??