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; }