# reverse linked list II

Reverse a linked list from positionmton. Do it in-place and in one-pass.

For example:\
Given`1->2->3->4->5->NULL`,m= 2 andn= 4,

return`1->4->3->2->5->NULL`.

**Note:**\
Givenm,nsatisfy the following condition:\
1 ≤m≤n≤ length of list.

分析

2种reverse方法，一种只改变中间指针指向，还有一种不断把next插入prev的后面，头插法？也是循环首位相接

```
ListNode temp = cur.next;
cur.next = prev;
prev = cur;
cur = temp;
```

```
头插法？不断把cur后面的Node插入prev的后面
ListNode move = cur.next;
cur.next = move.next;
move.next = prev.next;
prev.next = move;
```

此题从dummy开始，走m-1步到m前，然后开始头插法把后面的数不断插入m-1之后。

```
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prev = dummy, cur;
        for(int i = 0; i < m - 1; i ++){
            if(prev != null)
                prev = prev.next;
        }
        cur = prev.next;
        for(int i = m; i < n; i ++){
            ListNode move = cur.next;
            cur.next = move.next;
            move.next = prev.next;
            prev.next = move;
        }


        return dummy.next;
    }
}
```


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