python 栈的理解与使用

title： ①解决字符串的翻转

②堆和栈的区别？

 

>>> class Stack:
	def __init__(self):
		self.item = []
	def isEmpty(self):
		return len(self.item)==0
	def push(self,item):
		self.item.append(item)
	def pop(self):
		self.item.pop()
	def peek(self):
		if not self.isEmpty():
			return self.item[len(self.item)-1]
	def size(self):
		return len(self.item)

	
>>> s=Stack()
>>> type(s)
<class '__main__.Stack'>
>>> print(s)
<__main__.Stack object at 0x000002675BC61FD0>
>>> s.isEmpty()
True
>>> s.push(4)
>>> s.push('dog')
>>> s.peek()
'dog'
>>> s.pop()
>>> s.peek()
4
>>> s.isEmpty()
False
>>> s.size()
1
>>> 

class Node(object):
    def __init__(self, data, next = None):
        self.data = data
        self.next = next

class Stack(object):
    def __init__(self, top = None):
        self.top = top

    def push(self,data):
        #创建新的节点放到栈顶
        self.top = Node(data, self.top)

    def pop(self):
        #拿出栈顶元素，原来的栈发生改变
        if self.top is None:
            return None
        data = self.top.data
        self.top = self.top.next
        return data

    def peek(self):
        #查看栈顶元素，原来的栈不变
        return self.top.data if self.top is not None else None

    def isEmpty(self):
        return self.peek() is None


if __name__ == "__main__":
    stack = Stack()

stack = Stack()
print(type(stack))
stack1 = stack.push(1)
stack1 = stack.push(2)
print(stack.peek()) #查看栈顶元素
print("---------------")
stack.pop()
print(stack.peek())
print(stack.isEmpty())
print("dddddd")