Reverse Linked List II leetcode java

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.

题解：

经典的题目就是链表逆序啦，一般的链表逆序是让把链表从前到后都逆序，这个是给定了起始位置和结束位置，方法是一样的。

就是维护3个指针，startpoint，node1和node2。

startpoint永远指向需要开始reverse的点的前一个位置。

node1指向正序中第一个需要rever的node，node2指向正序中第二个需要reverse的node。

交换后，node1 在后，node2在前。这样整个链表就逆序好了。

代码如下：

1     public ListNode reverseBetween(ListNode head, int m, int n) {
2         ListNode newhead = new ListNode(-1);
3         newhead.next = head;
4
5         if(head==null||head.next==null)
6             return newhead.next;
7
8         ListNode startpoint = newhead;//startpoint指向需要开始reverse的前一个
9         ListNode node1 = null;//需要reverse到后面去的节点
10         ListNode node2 = null;//需要reverse到前面去的节点
11
12         for (int i = 0; i < n; i++) {
13             if (i < m-1){
14                 startpoint = startpoint.next;//找真正的startpoint
15             } else if (i == m-1) {//开始第一轮
16                 node1 = startpoint.next;
17                 node2 = node1.next;
18             }else {
19                 node1.next = node2.next;//node1交换到node2的后面
20                 node2.next = startpoint.next;//node2交换到最开始
21                 startpoint.next = node2;//node2作为新的点
22                 node2 = node1.next;//node2回归到node1的下一个，继续遍历
23             }
24         }
25         return newhead.next;
26     }

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