用 Python 实现堆栈

Question

我正在尝试使用数组使用 Python 实现一个简单的堆栈。我想知道是否有人可以让我知道我的代码有什么问题。

class myStack:
     def __init__(self):
         self = []

     def isEmpty(self):
         return self == []

     def push(self, item):
         self.append(item)

     def pop(self):
         return self.pop(0)

     def size(self):
         return len(self)

    s = myStack()
    s.push('1')
    s.push('2')
    print(s.pop())
    print s

Answer 1

堆栈是一个容器（线性集合），其中动态集合操作 按照后进先出（LIFO）原则进行。 只有一个指针——top，用于执行这些操作

使用数组实现堆栈的CLRS：

class Stack:
    """
    Last in first out (LIFO) stack implemented using array.
    """
    def __init__(self, capacity=4):
        """
        Initialize an empty stack array with default capacity of 4.
        """
        self.data = [None] * capacity
        self.capacity = capacity
        self.top  = -1

    def is_empty(self):
        """
        Return true if the size of stack is zero.
        """
        if self.top == -1:
            return True
        return False

    def push(self, element):
        """
        Add element to the top.
        """
        self.top += 1
        if self.top >= self.capacity:
            raise IndexError('Stack overflow!')
        else:
            self.data[self.top] = element

    def pop(self):
        """
        Return and remove element from the top.
        """
        if self.is_empty():
            raise Exception('Stack underflow!')
        else:
            stack_top = self.data[self.top]
            self.top -= 1
            return stack_top

    def peek(self):
        """
        Return element at the top.
        """
        if self.is_empty():
            raise Exception('Stack is empty.')
            return None
        return self.data[self.top]

    def size(self):
        """
        Return the number of items present.
        """
        return self.top + 1

测试实现：

def main():
    """
    Sanity test
    """
    stack = Stack()

    print('Size of the stack is:', stack.size())
    stack.push(3)
    print('Element at the top of the stack is: ', stack.peek())
    stack.push(901)
    print('Element at the top of the stack is: ', stack.peek())
    stack.push(43)
    print('Element at the top of the stack is: ', stack.peek())
    print('Size of the stack is:', stack.size())
    stack.push(89)
    print('Element at the top of the stack is: ', stack.peek())
    print('Size of the stack is:', stack.size())
    #stack.push(9)    # Raises IndexError
    stack.pop()
    print('Size of the stack is:', stack.size())
    stack.pop()
    print('Size of the stack is:', stack.size())
    stack.pop()
    print('Size of the stack is:', stack.size())
    print('Element at the top of the stack is: ', stack.peek())
    stack.pop()
    #print('Element at the top of the stack is: ', stack.peek())    # Raises empty stack exception

if __name__ == '__main__':
    main()

Answer 2

class Stack:
   s =[]

   def push(self, num):
       self.s.append(num)

   def pop(self):
       if len(self.s) == 0: # erro if you pop an empty list
           return -1
       self.s.remove(self.s[-1])

   def isEmpty(self):
       if len(self.s) == 0:
           return True
       else:
           return False

   def display(self): # this is to display how a stack actually looks like
       if self.isEmpty():
           print("Stack is Empty")
    
       for i in range(len(self.s)-1,-1,-1): # I haven't used reversed() since it will be obv
           print(self.s[i])


obj = Stack()
obj.push(3)
print(obj.isEmpty())
obj.push(4)
obj.display()
print("----")
obj.pop()
obj.display()

Answer 3

我想分享我的继承 Python List 的堆栈实现版本。我相信堆栈上的迭代应该以 LIFO 顺序发生。此外，应提供对pop-all() 的迭代以在弹出所有元素时进行迭代。我还添加了stack.clear() 来清空堆栈（就像我们在集合模块中的deque.clear() 中一样）。出于调试目的，我还覆盖了__repr__：

class Stack(list):

    def push(self, item):
        self.append(item)

    def top(self):
        return self[-1]

    def size(self):
        return len(self)

    def isempty(self):
        return self.size() == 0

    def __iter__(self):
        """ iter in lifo """
        return super(Stack, self).__reversed__()

    def __reversed__(self):
        return super(Stack, self).__iter__()

    def popall(self):
        try:
            while True:
                yield self.pop()
        except IndexError:
            pass

    def clear(self):
        del self[:]

    def __repr__(self):
        if not self:
            return '%s()' % self.__class__.__name__
        return '%s(%s)' % (self.__class__.__name__, super(Stack, self).__repr__())

这里是你如何使用它：

stack = Stack(range(5))
print "stack: ", stack  # stack:  Stack([0, 1, 2, 3, 4])

print "stack.pop() => ", stack.pop() # stack.pop() =>  4
print "stack.push(20) " # stack.push(20) 
stack.push(20)
for item in stack:
    print item  # prints 20, 3, 2... in newline
print "stack: ", stack # stack:  Stack([0, 1, 2, 3, 20])
print "stack pop all..."
for item in stack.popall(): # side effect to clear stack
    print item
print "stack: ", stack # stack:  Stack()

主要，我实现它是为了解决编程问题next greater element。

Answer 4

我在下面纠正了一些问题。此外，在抽象编程术语中，“堆栈”通常是您从顶部添加和删除的集合，但您实现它的方式是添加到顶部并从底部删除，这使其成为队列.

class myStack:
     def __init__(self):
         self.container = []  # You don't want to assign [] to self - when you do that, you're just assigning to a new local variable called `self`.  You want your stack to *have* a list, not *be* a list.

     def isEmpty(self):
         return self.size() == 0   # While there's nothing wrong with self.container == [], there is a builtin function for that purpose, so we may as well use it.  And while we're at it, it's often nice to use your own internal functions, so behavior is more consistent.

     def push(self, item):
         self.container.append(item)  # appending to the *container*, not the instance itself.

     def pop(self):
         return self.container.pop()  # pop from the container, this was fixed from the old version which was wrong

     def peek(self):
         if self.isEmpty():
             raise Exception("Stack empty!")
         return self.container[-1]  # View element at top of the stack

     def size(self):
         return len(self.container)  # length of the container

     def show(self):
         return self.container  # display the entire stack as list


s = myStack()
s.push('1')
s.push('2')
print(s.pop())
print(s.show())

Answer 5

下面是我的实现

class Stack:
    def __init__(self):
        self.items = list()
    def is_empty(self):
        return self.items == []
    def peek(self):
        if self.is_empty():
            print('Cannot peek empty stack')
            return
        else:
            return self.items[-1]
    def pop(self):
        if self.is_empty():
            print('Cannot pop an empty stack')
            return
        else:
            return self.items.pop()
    def size(self):
        return len(self.items)
    def push(self,data):
        self.items.append(data)

Answer 6

正确的实现应该包括__iter__，因为 Stack 需要是 LIFO 顺序。

class Stack:
    def __init__(self):
        self._a = []

    def push(self, item):
        self._a.append(item)

    def pop(self):
        return self._a.pop()

    def isEmpty(self):
        return len(self._a) == 0

    def __iter__(self):
        return reversed(self._a)

    def __str__(self):
        # return str(list(reversed(self._a)))
        return str(list(iter(self)))

def main():
    stack = Stack()
    stack.push('a')
    stack.push('b')
    stack.push('c')
    stack.pop()
    print(stack)
    if stack:
        print("stack not empty")
    stack.pop()
    stack.pop()
    if stack.isEmpty():
        print("stack empty")

if __name__ == '__main__':
    main()

Answer 7

Implementing a Stack in Python 来自《用算法和数据结构解决问题》一书中

Answer 8

下面是python中stack的简单实现。此外，它还返回任意时间点的中间元素。

  class Stack:
        def __init__(self):
            self.arrList = []

        def isEmpty(self):
            if len(self.arrList):
                return False
            else:
                return True

        def push(self, val):
            self.arrList.append(val)

        def pop(self):
            if not self.isEmpty():
                self.arrList[len(self.arrList)-1]
                self.arrList = self.arrList[:len(self.arrList)-1]
            else:
                print "Stack is empty"

        def returnMiddle(self):
            if not self.isEmpty():
                mid = len(self.arrList)/2
                return self.arrList[mid]
            else:
                print "Stack is empty"

        def listStack(self):
            print self.arrList

    s = Stack()
    s.push(5)
    s.push(6)
    s.listStack()
    print s.returnMiddle()
    s.pop()
    s.listStack()
    s.push(20)
    s.push(45)
    s.push(435)
    s.push(35)
    s.listStack()
    print s.returnMiddle()
    s.pop()
    s.listStack()

输出：

[5, 6]
6
[5]
[5, 20, 45, 435, 35]
45
[5, 20, 45, 435]

Answer 9

您的堆栈是一个数组...

class stacked(): # Nodes in the stack
    def __init__(self,obj,next):
        self.obj = obj
        self.next = next
    def getObj(self):
        return(self.obj)
    def getNext(self):
        return(self.next)

class stack(): # The stack itself
    def __init__(self):
        self.top=None
    def push(self,obj):
        self.top = stacked(obj,self.top)
    def pop(self):
        if(self.top == None):
            return(None)
        r = self.top.getObj()
        self.top = self.top.getNext()
        return(r)

Answer 10

我在评论中留下了http://docs.python.org/2/tutorial/datastructures.html#using-lists-as-stacks 的链接，但如果您想要一个自定义类型，可以为您提供push、pop、is_empty 和size 便捷方法，我只需子类list。

class Stack(list):
    def push(self, item):
        self.append(item)
    def size(self):
        return len(self)
    def is_empty(self):
        return not self

但是，正如我在 cmets 中所说，我可能会在这里直接使用 list，因为您实际上所做的只是给现有方法加上别名，这通常只会使您的代码更难在从长远来看，因为它需要人们使用它来学习您在原始界面之上的别名界面。

Answer 11

您的问题是您从列表的开头弹出，而您应该从列表的末尾弹出。堆栈是一种后进先出数据结构，这意味着当您从中弹出某些内容时，该内容将是您最后推送的内容。看看您的推送功能 - 它会将一个项目附加到列表中。这意味着它位于列表的末尾。但是，当您调用 .pop(0) 时，您将删除列表中的第一项，而不是您最后添加的一项。从 .pop(0) 中删除 0 应该可以解决您的问题。

Answer 12

分配给self 不会将您的对象变成一个列表（如果是这样，该对象将不再具有您所有的堆栈方法）。分配给self 只会改变一个局部变量。相反，设置一个属性：

def __init__(self):
    self.stack = []

并使用属性，而不是仅仅使用 self:

def push(self, item):
    self.stack.append(item)

另外，如果你想要一个堆栈，你想要pop() 而不是pop(0)。 pop(0) 会将您的数据结构变成一个（n 低效）队列。