LinkedList打印并将元素移动到另一个LinkedList

Question

首先从列表开始，我应该做LinkedList并将元素放入其中（我就是这么做的）。从该列表中，我必须将所有非字符串元素（整数）移动到另一个LinkedList并打印它们

初始化：

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

class LinkedList(object):
    def __init__(self):
        self.head = None
        self.tail = None
        self.size = 0

   def append(self,new_data):
        new_node = Node(new_data)
        if self.head is None:
            self.head = new_node
            return
        last = self.head
        while(last.next):
            last = last.next
        last.next = new_node 
 
def Lista(lista):
    linked = LinkedList()
    for element in lista:
        linked.append(element)
    return linked

打印函数和将元素移动到另一个linkedlist的函数（izdvoji）：

def ispis(self):
    temp = self.head
    while (temp):
        print (temp.data)
        temp = temp.next
    print("------")

def izdvoji(self):
    linked = LinkedList()
    temp = self.head
    while(temp):
        if isinstance (temp.data, str):
            linked.append(temp.data)
            temp = temp.next //edited
        else:
            temp = temp.next

if __name__ == '__main__':
    L = Lista([33, "asp","oop1",5,21,"python",2,"c++"])
    ispis(izdvoji(L))
    ispis(L)

结果必须是：

ASP、OOP1、Python、C++ +P>

33,5,21,2

抱歉，代码太长，但我想让你看看

编辑功能：

def izdvoji(linkedLista):
    linked = LinkedList()
    temp = linkedLista.head
    while(temp != None):
        if isinstance (temp.data, str):
            linked.append(temp.data)
        temp = temp.next
   
    temp = linkedLista.head
    while(temp != None):
        if temp == linkedLista.head and isinstance (temp.data, str):
            linkedLista.head = temp.next
            print(temp.data)
    
        if temp.next != None and isinstance (temp.next.data, str): 
            temp.next = temp.next.next
           
        temp = temp.next
    return linked

输出：

ASP、OOP1、Python、C++、

---

33,1,5,21,2,

Answer 1

print函数接受默认为"\n"的end参数。您可以使用", "打印列表

要将节点从一个列表移动到另一个列表，需要执行两个步骤

1. 将节点复制到新列表
2. 从旧列表中删除节点

您没有一个delete方法来处理这个问题，所以我必须自己编写delete方法

链表的删除方法

    def delete(self, node_data):
        temp = self.head

        # If list is empty
        if temp is None:
            return

        # If the head need to be removed
        if temp.data == node_data:
            self.head = temp.next
            return

        # Iterate over the list and stops when the next node is our required node
        while temp is not None and temp.next.data != node_data:
            temp = temp.next

        # if temp is none that means we reached the end of list
        if temp is None:
            return

        # removing the node
        temp.next = temp.next.next

def ispis(linked_list):
    temp = linked_list.head
    while (temp):
        print(temp.data, end=", ")
        temp = temp.next
    print("\n   ")

输出：

33, asp, oop1, 5, 21, python, 2, c++, 
   

就复制而言。如果条件为true，则忘记更新临时值。从技术上讲，这里也不需要其他东西。移动未执行的元素后，还需要返回列表

def izdvoji(linked_list):
    linked = LinkedList()
    temp = linked_list.head
    while temp:
        if isinstance(temp.data, str):
            linked.append(temp.data) # Copy node to new list
            linked_list.delete(temp.data) # delete from old list
        temp = temp.next
    return linked

输出：

asp, oop1, python, c++, 
   
33, 5, 21, 2, 
   

主代码：

if __name__ == '__main__':
    L = Lista([33, "asp","oop1",5,21,"python",2,"c++"])
    ispis(izdvoji(L))
    ispis(L)

注意：按照约定，self仅用于引用类的属性和方法izdvoji和ispis不是类方法。所以在这里使用self是没有意义的