Reverse Linked List II - LeetCode

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.

反转单链表的精髓就是，每次把最后一个Node提到首部的后方，就是把最后一个Node提到最前面。

注意一定要设置一个“哨兵头结点”，以保证第一个结点也能被反转。

ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;

题目AC的代码如下：

public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (m >= n || head == null) {
            return head;
        }
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;
        
        for (int i = 1; i < m; i++) {
            if (head == null) {
                return null;
            }
            head = head.next;
        }
        
        ListNode one = head;
        ListNode two = head.next;
        ListNode three = two.next;
        ListNode temp = three.next;
        for(int j = m; j < n ; j++){
            temp = three.next;
            three.next = one.next;
            one.next = three;
            two.next = temp;
            three = two.next;
            
        }
        
         return dummy.next;

    }
}