92. Reverse Linked List II
1. 原始题目
Reverse a linked list from position m to n. Do it in one-pass.
Note: 1 ≤ m ≤ n ≤ length of list.
Example:
Input: 1->2->3->4->5->NULL, m = 2, n = 4 Output: 1->4->3->2->5->NULL
2. 题目理解
将m至n的链表反转。注意下标这里在链表中实际上是m-1至n-1。
我的思路是先遍历m步，然后在n-m+1遍历过程中将结点反转。然后将这一整段链表重新链接即可。这样做一个缺点就是需要多个标记点。
3. 解题
1 # Definition for singly-linked list. 2 # class ListNode: 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 7 class Solution: 8 def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode: 9 if not head.next: 10 return head 11 m = m-1 # 修正下标 12 n = n-1 13 dummy = ListNode(0) 14 dummy.next = head 15 record = dummy 16 17 for _ in range(m): # 先走到需要反转的地方 18 dummy = dummy.next 19 20 end = dummy.next # 记录断点 21 i = dummy 22 j = i.next # i，j，k 负责反转链表 23 if j: 24 k = j.next 25 for _ in range(n-m+1): 26 if j: 27 j.next = i 28 i = j 29 j = k 30 if k: 31 k = k.next 32 else: 33 return head 34 dummy.next = i # 重新链接链表一端 35 end.next = j # 重新连接链表另一端 36 37 return record.next
验证：
1 # 新建链表1 2 listnode1 = ListNode_handle(None) 3 s1 = [1,2,3,44,555,66,7,8] 4 for i in s1: 5 listnode1.add(i) 6 7 listnode1.print_node(listnode1.head) 8 9 s = Solution() 10 head = s.reverseBetween(listnode1.head,4,5) 11 listnode1.print_node(head)
1 2 3 44 555 66 7 8
1 2 3 555 44 66 7 8
LeetCode上有另一种解题思路和我的极其类似：这里粘过来对比学习一下：
1 class Solution(object): 2 def reverseBetween(self, head, m, n): 3 """ 4 :type head: ListNode 5 :type m: int 6 :type n: int 7 :rtype: ListNode 8 """ 9 if not head or m == n: return head 10 p = dummy = ListNode(None) 11 dummy.next = head 12 for i in range(m-1): p = p.next 13 tail = p.next 14 15 for i in range(n-m): 16 tmp = p.next # a) 17 p.next = tail.next # b) 18 tail.next = tail.next.next # c) 19 p.next.next = tmp # d) 20 return dummy.next
流程如下：
