题解 | #链表内指定区间反转#

public ListNode reverseBetween2(ListNode head, int m, int n) {
        // write code here
        if (m == n) {
            return head;
        }
        if (head == null || head.next == null) {
            return head;
        }
        //1.记录m-1位置节点l
        ListNode cur = head;
        ListNode lNode = null;
        //1.1当m=1时，直接反转1至n的链表，此时lNode为空。
        for (int i = 1; i < m; i++) {
            if (i == m - 1) {
                lNode = cur;
            }
            cur = cur.next;
        }
        //2.记录反转链表的尾节点m,记录反转节点的头节点n,记录n.next节点o
        ListNode mNode = cur;
        ListNode pre = null;//n
        ListNode next = null;//o
        for (int i = m; i <= n; i++) {
            next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }
        //3.连接l->n,m->o
        //1.2 如果lNode为空的话，直接将pre就是头节点
        if (m > 1) {
            lNode.next = pre;
        } else {
            head = pre;
        }
        mNode.next = next;
        return head;
    }