leetcode 用队列模拟栈

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 None
        for 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 None
        for 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  temp

    def 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 None
        for 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 None
        for i in range (len(self.que)-1):
            self.que.append(self.que.popleft())
        last= self.que.popleft()
        self.que.append(last)
        return last
    def 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 temp

    def 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 temp

    def 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 temp

    def 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 temp

    def 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()