# [LeetCode 92] Reverse Linked List II

### 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 ≤ mn ≤ 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) {
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;