Reverse Link List II

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)

/**
 * Definition for singly-linked list.
 * 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;
    }
}