如何检查项目是否在链接列表中？

Question

基本上，我想检查一个项目是否在链接列表中。该函数被描述为__contains__，其中如果输入3 in myList，则将返回True或False，具体取决于链接列表中是否存在整数3。

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

    def __str__(self): 
     return str(self.item) 

class LinkedList: 
    def __init__(self): 
     self.head = None 
     self.count = 0 

    def __str__(self): 
     current = self.head 
     ans = str(current) 
     for _ in range(len(self)): 
      current = current.next 
      ans += '\n' 
      ans += str(current) 
     return ans 

    def _get_node(self,index): 
     if 0<= index< len(self): 
      current = self.head 
      while index>0: 
       current = current.next 
       index -=1 
      return current 

    def __contains__(self,item):   #need some help here 
     if self.isEmpty(): 
      raise StopIteration("List is empty") 
     if self.head == item: 
      return True 
     nextItem = self.head.next 

    def insert(self,index,item): 
     if index < 0 or index > len(self): 
      raise IndexError("Index is out of range") 
     else: 
      newNode = Node(item) 
      if index == 0: 
       newNode.next = self.head 
       self.head = newNode 
      else: 
       before = self._get_node(index-1) 
       newNode.next = before.next 
       before.next = newNode 
      self.count+=1 
      return True 

if __name__ == "__main__": 
    L = LinkedList() 
    L.insert(0, 0) 
    L.insert(1, 1) 
    L.insert(2, 2) 
    L.insert(3, 3) 
    print(0 in L) 

在迭代链表和检查项目是否在其中时，我感到非常困惑。最后一行中的print(0 in L)应该返回True，因为0确实在链接列表中。

Answer 1

这里是你的问题的答案：

def __contains__(head,data): 
    if head == None: 
     return False 
    else: 
     p = head 
     while p is not None: 
      if p.data == data: 
       return True 
      p = p.next 
     return False