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.

思路很简单：设置两个指针：fast往前移动n步，low往前移动m步，然后依次将low和fast之间的节点挨个插到fast的后面即可。注意：为去除头结点的特殊性，需要用到虚拟头结点技术。

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