Sword finger offer (V): queue with two stacks

solution 1： Initialize two empty stacks , When stack2 When is not empty , stay stack2 The top element in the stack is the first element to enter the queue , It can pop up . If stack2 It's empty time , We put stack1 The elements in pop up one by one and press in stack2. Because the element that enters the queue first is pressed to stack1 At the bottom of the stack , After ejecting and pressing in, it is in stack2 Top of stack , You can pop up directly . If there are new elements d Insert , We press it directly into stack1 that will do ！

class Solution:
    def __init__(self):
        self.stack1 = []
        self.stack2 = []
    def push(self, node):
        # write code here
        return self.stack1.append(node)
    def pop(self):
        # return xx
        if self.stack2 == []:
            while self.stack1:
                self.stack2.append(self.stack1.pop())
            return self.stack2.pop()
        return self.stack2.pop()