Leetcode -- Reverse Linked List II select some nodes in the Linked List in Reverse order (AC)

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given1->2->3->4->5->NULL, M = 2 and n = 4,

Return1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

It is still complicated to deal with this problem. Many boundary test cases need to be considered. My general idea is to mark the previous node of m and the node behind n cyclically, and use the node behind n as the pre of the current reverse node first, then, complete the reverse order of the selected node parts n-m times, and point the previous node of the marked m node to the header node of the backward part. Consider special cases. The code is as follows:

class Solution {public:    ListNode *reverseBetween(ListNode *head, int m, int n) {        if(head==NULL || m<0 || n<0)            return head;        if(head->next == NULL || m==n)            return head;        ListNode *head2=NULL,*pre,*cur,*temp=head;        for(int i=0; i
 
  next;        }        for(int i=m;i
  
   next;            cur->next = pre;            pre = cur;            cur = temp;        }        if(m==1)            return pre;        head2->next = pre;        return head;    }};