What is the difference between pop() and pop()[0]?-CodePudding

I am leetcoding and implementing a stack when I saw the answer use a pop function using self.items.pop()[0].

def pop(self):
    if len(self.items) > 0:
        return self.items.pop()[0]

In this case does this simply do the same thing pop() does?

I understand that pop(x) will pop the value at the xth index. I would appreciate some examples of pop()[x].

CodePudding user response：

I'm assuming, based on your phrasing, that self.items is a list. The pop method for lists, as you've pointed out, can be used to remove and return an element of the list at a specified index. If you omit the index, then the last item is removed and returned.

Either way, pop returns the item that was removed. If you put [0] after the call to pop, you'll get the first item of the item that was popped. So, if self.items is [[1, 2, 3], [4, 5, 6]], then self.items.pop() would be [4, 5, 6] and therefore self.items.pop()[0] would be 4.

CodePudding user response：

You can try it.

def pop(self):
if len(self.items) > 0:
    return self.items[1].pop(0)

Her items[1] is items[items list index]

items = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]