Reverse Linked List II

题目：

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.

    public ListNode reverseBetween(ListNode head, int m, int n) {
        if(head == null) return null;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        ListNode start = head;
        
        for (int i = 1; i < m; i++) {
            start = start.next;
            pre = pre.next;
        }
        ListNode end = start;
        ListNode then = start.next;
        ListNode temp = then;
        for (int j = 0; j < n - m;  j++) {
            temp = then.next;
            then.next = start;
            start = then;
            then = temp;
        }
        pre.next = start;
        end.next = then;
        return dummy.next;
    }

感觉写的挺丑的。。。