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.
Note 2:
Usually the version often seen in the interviews is reversing the whole linked list which is obviously an easier version of this question
Method:

As we traverse the list, we keep track of the key nodes, and connect them with each other.

Solution:

Time: O(n)
Space: o(1)

/**
* class ListNode {
*     public int val;
*     public ListNode next;
*     ListNode(int x) { val = x; next = null; }
* }
*/
public class Solution {
//    2           4
// 1->2->3->4->5->NULL
// p   l          m
// 1->2<-3<-4->5->NULL
// 1->4->3->2->5->NULL
public ListNode reverseBetween(ListNode A, int B, int C) {
ListNode dummy = new ListNode(0);
dummy.next = A;
ListNode p = dummy;
ListNode curr = A;
int n = 1;
while (n < B && curr != null) {
p = curr;
curr = curr.next;
n ++;
}
ListNode l = curr;
ListNode prev = null;
while (n <= C && curr != null) {
ListNode next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
n ++;
}
ListNode m = prev;
p.next = m;
l.next = curr;
return dummy.next;
}
}