描述
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->nullptr, m = 2 and n = 4,
return 1->4->3->2->5->nullptr.
Note: Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of list.
分析

這題非常繁瑣，有很多邊界檢查，15 分鐘內做到 bug free 很有難度！

代碼

 1 public class ReverseLinkedList {
 2 
 3     public static void main(String[] args) {
 4         // TODO Auto-generated method stub
 5         ListNode head = new ListNode(1);
 6         // head.add(1);
 7         head.add(2);
 8         head.add(3);
 9         head.add(4);
10         head.add(5);
 
11         head.print();
12         System.out.println();
13         reverseBetween(head, 2, 4).print();
14     }
15 
16     public static ListNode reverseBetween(ListNode head, int m, int n) {
17         ListNode newhead = new ListNode(-1);
18         newhead.next = head;
19         if (head == null
 
 || head.next == null)
20             return newhead.next;
21 
22         ListNode startpoint = newhead;
23         ListNode node1 = null;
24         ListNode node2 = null;
25         for (int i = 0; i < n; i++) {
26             if (i < m - 1) {
27                 startpoint = startpoint.next;// 尋找真正的startpoint m-1的位置
28             } else if (i == m - 1) { // 開始變換的第一輪，找到初始位置
29                 node1 = startpoint.next;
30                 node2 = node1.next;
31             } else {
32                 node1.next = node2.next; // node1交換到node2後面
33                 node2.next = startpoint.next; // node2連startpoint後面的數，每次都移到最前
34                 startpoint.next = node2; // startpoint連node2
35                 node2 = node1.next; // node2移到node1後面，進行下一次的交換
36 
37             }
38         }
39         return newhead.next;
40     }
41 }

Reverse Linked List II