def deleteDuplication(pHead):
head = LinkedList.ListNode()
head.next = pHead
# 以后使用p代表前一个节点,q代表后一个节点
# 当只需要一个节点操作，如遍历时，使用p
# 当需要执行诸如删除操作时，使用p和q，删除q所指的节点就是讲p.next置为q.next
p = head
q = pHead
# 遍历中，判断q是否为None即可
# 由于这里需要比较q.next，所以判断q.next是否None即可
while q.next: #r
if q.val != q.next.val:
if q != p.next:
p.next = q.next
else:
p = q
q = q.next
# 判断链表结尾是重复的问题
if q != p.next:
p.next = q.next
return head.next
def merge_two_sorted_list(list1,list2):
# Create placeholder for result
dummy_head = tail = ll.ListNode()
while list1 and list2:
if list1.data < list2.data:
tail.next_node, list1 = list1,list1.next_node
else:
tail.next_node, list2 = list2, list2.next_node
tail = tail.next_node
tail.next_node = list1 or list2
return dummy_head.next_node
def reverse_sublist(linked_list, s, f):
dummy_head = sublist_head = ll.ListNode(0, linked_list)
# Sets sublist_head to node before s
for _ in range(1, s):
sublist_head = sublist_head.next_node
# Sets sublist_itter to Node at s
sublist_itter = sublist_head.next_node
for _ in range(f - s):
temp = sublist_itter.next_node
sublist_itter.next_node = temp.next_node
temp.next_node = sublist_head.next_node
sublist_head.next_node = temp
return dummy_head.next_node
def revrese_linked_list(head):
nodes = []
while head:
nodes.append(head)
head = head.next_node
# Reset next_node to None to prevent last item looping on itself
for node in nodes:
node.next_node = None
new_head = dummy_head = ll.ListNode(0)
nodes = nodes[::-1]
new_head.next_node = nodes[0]
for i in range(len(nodes) - 1):
nodes[i].next_node = nodes[i + 1]
return dummy_head.next_node
