leetcode92~Reverse Linked List II

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.

根本还是单链表的翻转，只不过需要预先保存好一些节点的值

public class ReverseLinkedListII {
    public ListNode reverseBetween(ListNode head, int m, int n) {

        if(head==null || m>n) {
            return null;
        }

        //设置一个dummyNode 便于操作
        ListNode dummyNode = new ListNode(-1);
        dummyNode.next = head;
        head = dummyNode;
        //到m-n区间前面一个节点
        for(int i=1;i<m;i++) {
            if(head==null) {
                return null;
            }
            head = head.next;
        }

        //保存n-m区间前一个节点
        ListNode pre = head;
        //保存n-m区间第一个节点
        ListNode fNode = head.next;
        ListNode lastNode = null;
        ListNode curNode= fNode;
        for(int i=m;i<=n;i++) {
            ListNode tmp = curNode.next;
            curNode.next = lastNode;
            lastNode = curNode;
            curNode = tmp;
        }
        pre.next = lastNode;
        fNode.next = curNode;

        return dummyNode.next;

    }
}