Reverse Linked List II

leetcode 92.

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.

Solution. Should return head if head is null or head only has 1 element.

We use a dummyHead pointing to head. Then we put start to the m – 1 position, put pre to m position, put curr to m + 1 position. Each time, we move curr after start. At the beginning, start points to dummyHead.  Process for reversing [2..4] is like below.

If there are only 2 elements, the initial is like this:

public static ListNode reverseBetween(ListNode head, int m, int n) {
    if (head == null || head.next == null) {
        return head;
    }
    ListNode dummyHead = new ListNode(-1);
    dummyHead.next = head;
    ListNode start = dummyHead;
    for (int i = 1; i < m; i++) {
        start = start.next;
    }
    ListNode pre = start.next, curr = start.next.next;
    for (int i = 1; i <= n - m; i++) {
        pre.next = curr.next;
        curr.next = start.next;
        start.next = curr;
        curr = pre.next;
    }
    return dummyHead.next;
}

Check my code on github.