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.

题意：反转链表的指定一段。

思路：这道题是Reverse Linked List的followup，把反转整个链表限定为反转指定一段。可以把反转的解法套在指定范围上，反转之后还要把它和原来的两端连接起来，所以需要记录一下两端连接的节点，以及反转后的这段链表的头尾节点。

public ListNode reverseBetween(ListNode head, int m, int n) {
    if (head == null || head.next == null || m < 0 || n <= m) {
        return head;
    }

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

    //连接反转段的端点
    ListNode left = root;
    ListNode right = new ListNode(0);

    //反转段的左右端点
    ListNode rLeft = new ListNode(0);
    ListNode rRight = new ListNode(0);

    //反转需要的两个指针
    ListNode pre = new ListNode(0);
    ListNode cur = new ListNode(0);

    for (int i = 0; i < m - 1; i++) {
        left = left.next;
    }

    rRight = left.next;
    pre = left.next;
    cur = pre.next;

    for (int i = 0; i < n - m; i++) {
        ListNode next = cur.next;
        cur.next = pre;
        pre = cur;
        cur = next;
    }
    rLeft = pre;
    right = cur;

    left.next = rLeft;
    rRight.next = right;

    return root.next;
}