Python 使用linkedlist查找数字的频率
使用linkedlist查找数字的频率 运行以下代码时,获取SIGTSTP-时间限制超出错误。谁能帮我一下我哪里弄错了Python 使用linkedlist查找数字的频率,python,python-3.x,linked-list,Python,Python 3.x,Linked List,使用linkedlist查找数字的频率 运行以下代码时,获取SIGTSTP-时间限制超出错误。谁能帮我一下我哪里弄错了 class Element(object): def __init__(self,value): self.value = value self.next = None class LinkedList(object): def __init__(self, head = None): self.head = h
class Element(object):
def __init__(self,value):
self.value = value
self.next = None
class LinkedList(object):
def __init__(self, head = None):
self.head = head
def append(self, new):
current = self.head
if self.head:
while current.next:
current = current.next
current.next = new
else:
self.head = new
def traverse(self):
current = self.head
while current != None:
print(current.value)
current = current.next
arr = list(map(int, input().split()))
ll = LinkedList()
for i in arr:
e = Element(i)
ll.append(e)
ll.traverse()
def frequency(a):
current = a.head
while current != None:
count = 1
while current.next != None:
if current.value == current.next.value:
current+=1
if current.next.next != None:
current.next = current.next.next
else:
current.next = None
print(str(current.value)+" : " + str(count))
current = current.next
frequency(ll)
除了频率之外,一切看起来都很好。您需要保留两个引用,一个指向当前元素,另一个将从当前元素开始遍历列表的其余部分。这能让你继续下去吗
还要注意,您当前的实现将修改底层链表,而您确实可以使用指针进行“跳过”,以防止多次列出同一元素,我建议避免以这种方式修改基础结构。为什么要在链接列表中添加项目的值?@FarhoodET我正在将元素对象添加到链接列表中