leetcode：232. Realize queue with stack [two stacks, one auxiliary and one simulated queue]

analysis

st1 Keeping the top of the stack is opposite to the top , In this case, it needs to be reversed twice
That is to say st1 Accept a new one at a time x
Pour it all in first st2, then st2 Again append x
Finally, put st2 Pour it all in st1

ac code

class MyQueue:

    def __init__(self):
        self.st1 = [] #  From the top of the stack to the bottom of the stack is from the head of the queue to the end of the queue 
        self.st2 = [] #  Auxiliary stack 


    def push(self, x: int) -> None:
        if not self.st1:
            self.st1.append(x)
        else:
            #  hold st1 Come back 
            while self.st1:
                self.st2.append(self.st1.pop())
            self.st2.append(x) #  To the end 
            #  Back again st1, become queue
            while self.st2:
                self.st1.append(self.st2.pop())
        #print(len(self.st1))


    def pop(self) -> int:
        #print(len(self.st1))
        return self.st1.pop()
        


    def peek(self) -> int:
        return self.st1[-1]


    def empty(self) -> bool:
        
        return len(self.st1) == 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()

summary

Double stack implementation queue