### Question

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.

### Stats

Frequency | 2 |

Difficulty | 3 |

Adjusted Difficulty | 3 |

Time to use | ---------- |

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

### Analysis

**This is a very standard interview question of LinkedList**.

### Solution

**Keep 2 pointers, one of which points to preReversePosition, another one points to finalTailOfTheReversedPart**. Each time, I will get next element and insert it between the 2 pointers mentioned above. See below image for details:

The coding isn’t easy, there can be a lot of details being omitted. Just think of it in this way: in the above picture, 123 is broken away from 45678. Each time, get next element from 45678, and insert it right after 2. Do this 2 times, so that 4,5 are moved. In the end, make 3 point to 6, and solution done.

### Code

**First, my code**

```
public ListNode reverseBetween(ListNode head, int m, int n) {
ListNode preHead = new ListNode(0);
preHead.next = head;
ListNode before = preHead;
for (int i = 1; i < m; i ++)
before = before.next;
ListNode rTail = before.next;
ListNode cur = before.next.next;
for (int i = 0; i < n - m; i ++) {
ListNode temp = cur.next;
cur.next = before.next;
before.next = cur;
cur = temp;
}
rTail.next = cur;
return preHead.next;
}
```

**Second, **

A lot of people have similar solutions, so I won’t post any of their code. Reading it won’t help, write it yourself is actually important.

A lot of people have complained about this problem not easy to write.