Leetcode [linked list]: reverse Linked List II

Reverse a linked list from position m to N. Do it in-place and in one-pass.
For example:
Given1->2->3->4->5->NULL, M = 2 and N = 4,
Return1->4->3->2->5->NULL.
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

The idea is simple: set two pointers: fast move n steps forward, low move M steps forward, and then insert the nodes between low and fast one by one to the end of fast. Note: to remove the special characteristics of the header node, you need to use the virtual header node technology.

ListNode *reverseBetween(ListNode *head, int m, int n) {    ListNode *dummyHead = new ListNode(0);    dummyHead->next = head;    ListNode *fast = head;    for (int i = 0; i < n - 1; ++i)        fast = fast->next;    ListNode *low  = dummyHead;    for (int i = 0; i < m - 1; ++i)        low  = low ->next;    for (int i = 0; i < n - m; ++i) {        ListNode *curr = low->next;        low ->next = low ->next->next;        curr->next = fast->next;        fast->next = curr;    }    head = dummyHead->next;    delete dummyHead;    return head;}

Leetcode [linked list]: reverse Linked List II