链表删除系列 Linked List Delete Series
Idea
- Find the node to be removed
cur.next = cur.next.next
237. Delete Node in a Linked List
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def deleteNode(self, node):
"""
:type node: ListNode
:rtype: void Do not return anything, modify node in-place instead.
"""
cur = node
while cur and cur.next:
cur.val = cur.next.val
if cur.next.next == None:
cur.next = None
else:
cur = cur.next
19. Remove Nth Node From End of List
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
dummy = ListNode(next=head)
slow = dummy
fast = dummy
for i in range(n):
fast = fast.next
while fast.next:
slow = slow.next
fast = fast.next
slow.next = slow.next.next
return dummy.next
83. Remove Duplicates from Sorted List
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
dummy = ListNode(next=head)
cur = dummy.next
while cur and cur.next:
if cur.val == cur.next.val:
cur.next = cur.next.next
else:
cur = cur.next
return head
82. Remove Duplicates from Sorted List II
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
dummy = ListNode(next=head)
cur = dummy
while cur.next and cur.next.next:
val = cur.next.val
if val == cur.next.next.val:
while cur.next and val == cur.next.val:
cur.next = cur.next.next
else:
cur = cur.next
return dummy.next