[LeetCode] Reverse Linked List II

Problem:

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

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

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

Solution:O(n)

public class Solution {
   public ListNode reverseBetween(ListNode head, int m, int n) {
      if(head==null || head.next == null) 

         return head;
        
       ListNode pre = new ListNode(0);
       pre.next=head;
       head=pre;
       ListNode n1=head;
       int k=m-1;
       while(k>0){
          n1=n1.next;
          k--;
       }
        
       pre=n1;
       n1=n1.next;
       k=n-m;
       while(n1.next!=null && k>0){
          ListNode tmp =n1.next;
          n1.next=tmp.next;
          tmp.next=pre.next;
          pre.next=tmp;
          k--;
       }
       return head.next;
    }
}

Key Concept: pre ... n1 tmp