学院路网站建设乐陵森源木业全屋定制
这个其实只需要一个队列就可以的，但是我这里用的是2个队列进行替换， 先转n-1个到空的队列，
然后在此基础上进行队列的互换，把剩下的那一个元素所在的队列进行poleft操作就可以了。
class MyStack:def __init__(self):self.q1_in= deque()self.q2_out =deque()def push(self, x: int) -> None:self.q1_in.append(x)def pop(self) -> int:if self.empty():return Nonefor i in range(len(self.q1_in)-1):self.q2_out.append(self.q1_in.popleft())self.q1_in,self.q2_out =self.q2_out,self.q1_in return self.q2_out.popleft()def top(self) -> int:if self.empty():return Nonefor i in range(len(self.q1_in)-1):self.q2_out.append(self.q1_in.popleft())self.q1_in,self.q2_out =self.q2_out,self.q1_in temp=self.q2_out.popleft()self.q1_in.append(temp)return tempdef empty(self) -> bool:return len(self.q1_in) == 0 and len(self.q2_out) == 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()用一个que的方式
class MyStack:def __init__(self):self.que=deque()def push(self,x:int)-> None:self.que.append(x)def pop(self)->int:if self.empty():return Nonefor i in range (len(self.que)-1):self.que.append(self.que.popleft())return self.que.popleft()def top(self)-> int:if self.empty():return Nonefor i in range (len(self.que)-1):self.que.append(self.que.popleft())last= self.que.popleft()self.que.append(last)return lastdef empty(self)->bool:return not self.que用栈实现队列
232. 用栈实现队列 - 力扣（LeetCode）
class MyQueue:def __init__(self):self.stack1=[]self.stack2=[]def push(self, x: int) -> None:self.stack1.append(x)def pop(self) -> int:for _ in range(len(self.stack1)):self.stack2.append(self.stack1.pop())temp =self.stack2.pop()#注意这里需要把stack2 替换回原来的stack1的方式不能直接用替换 s1,s2 = s2,s1 ,这样下一次的pop 就不满足 队列的先进先出 for _ in range (len(self.stack2)):self.stack1.append(self.stack2.pop())return tempdef peek(self) -> int:for _ in range(len(self.stack1)):self.stack2.append(self.stack1.pop())temp= self.stack2.pop()self.stack2.append(temp) for _ in range (len(self.stack2)):self.stack1.append(self.stack2.pop())return tempdef empty(self) -> bool:return len(self.stack1)==0# Your MyQueue object will be instantiated and called as such:
# obj = MyQueue()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.peek()
# param_4 = obj.empty()用代码随想录的思想， 把 stack——in 和stack_out 看成一个整体队列， 这个时候 stack——out不不是一个中转的容器，特别在peek中我们要把pop出去的元素继续添加回stack_out .
class MyQueue:def __init__(self):self.stack1=[]self.stack2=[]def push(self, x: int) -> None:self.stack1.append(x)def pop(self) -> int:for _ in range(len(self.stack1)):self.stack2.append(self.stack1.pop())temp =self.stack2.pop()#注意这里需要把stack2 替换回原来的stack1的方式不能直接用替换 s1,s2 = s2,s1 ,这样下一次的pop 就不满足 队列的先进先出 for _ in range (len(self.stack2)):self.stack1.append(self.stack2.pop())return tempdef peek(self) -> int:for _ in range(len(self.stack1)):self.stack2.append(self.stack1.pop())temp= self.stack2.pop()self.stack2.append(temp) for _ in range (len(self.stack2)):self.stack1.append(self.stack2.pop())return tempdef empty(self) -> bool:return len(self.stack1)==0# Your MyQueue object will be instantiated and called as such:
# obj = MyQueue()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.peek()
# param_4 = obj.empty()