基本上,我想检查项目是否在链接列表中。该函数被描述为__contains__
,如果我键入3 in myList
,它将返回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)
当涉及迭代链表并检查项目是否在其中时,我很困惑。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。
答案 0 :(得分:1)
以下是您的问题的答案:
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