一、反转链表
LeetCode 206题:https://leetcode.cn/problems/reverse-linked-list/description/
给你单链表的头节点 head
,请你反转链表,并返回反转后的链表。
解析:
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
cur, pre = head, None
while cur:
tmp = cur.next # 暂存后继节点 cur.next
cur.next = pre # 修改 next 引用指向
pre = cur # pre 暂存 cur
cur = tmp # cur 访问下一节点
return pre
二、反转指定区间链表
LeetCode 92题:https://leetcode.cn/problems/reverse-linked-list-ii/description/?envType=study-plan-v2&envId=top-interview-150
给你单链表的头指针 head
和两个整数 left
和 right
,其中 left <= right
。请你反转从位置 left
到位置 right
的链表节点,返回 反转后的链表 。
解析:
class Solution:
def reverseBetween(self, head: Optional[ListNode], left: int, right: int) -> Optional[ListNode]:
dummy = ListNode(next = head)
p0 = dummy
for _ in range(left - 1):
p0 = p0.next
pre = None
cur = p0.next
for _ in range(right - left + 1):
tmp = cur.next
cur.next = pre
pre = cur
cur = tmp
p0.next.next = cur
p0.next = pre
return dummy.next
三、k个一组反转链表
LeetCode 第25题:https://leetcode.cn/problems/reverse-nodes-in-k-group/description/
给你链表的头节点 head
,每 k
个节点一组进行翻转,请你返回修改后的链表。
k
是一个正整数,它的值小于或等于链表的长度。如果节点总数不是 k
的整数倍,那么请将最后剩余的节点保持原有顺序。
你不能只是单纯的改变节点内部的值,而是需要实际进行节点交换。
class Solution:
def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
n = 0
cur = head
while cur:
n += 1
cur = cur.next
dummy = ListNode(next = head)
p0 = dummy
pre = None
cur = p0.next
while n >= k :
n -= k
for _ in range(k):
tmp = cur.next
cur.next = pre
pre = cur
cur = tmp
tmp = p0.next
p0.next.next = cur
p0.next = pre
p0 = tmp
return dummy.next