Sword finger offer (21): push in and pop-up sequence of stack

solution 1： Borrow an auxiliary stack , Traversing the stack order , First put the first one on the stack , Here is 1, It then determines whether the top element on the stack is the first element in the stack order , Here is 4, Obviously 1！=4, So we continue to stack , Until the stack is equal , Push an element , Moves the stack order one bit backward , Until it's not equal , So the loop isopressure order traversal completes , If the secondary stack is not already empty , Indicates that the pop-up sequence is not the pop-up order of the stack ！

class Solution:
    def IsPopOrder(self, pushV, popV):
        # write code here
        if len(popV) == 0 or len(pushV) != len(popV):
            return False
        stackData = []
        for i in pushV:
            stackData.append(i)# Borrow auxiliary stack , Traversing the stack order ！
            while len(stackData) and stackData[-1] == popV[0]:
                stackData.pop()
                popV.pop(0)
        if len(stackData):
            return False
        return True