反转链表

    反转链表是一个常见的面试题，现在出现了N多种的变形，各种增加难度。今天看到了一个从m反转到n的，于是手写代码试了试。发现要想把所有情况覆盖全达到bug free还是挺难的。于是，根据各种反转链表题总结了一些比较简练的代码。比较：接口限定的反转链表。

// From curr reverse to end, return reversed linked list head.
ListNode *unguarded_reverse(ListNode *prev, ListNode *curr, ListNode *end)
{
    ListNode *next = NULL;

    while (curr != end)
    {
        next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }

    return prev;
}

// Reverse linked list from m to n, return head.
// For example:
// Given 1->2->3->4->5->NULL, m = 2 and n = 4,
// return 1->4->3->2->5->NULL.
ListNode *reverse_between(ListNode *head, int m, int n)
{
    if (head == NULL || m <= 0 || n <= 0)
    {
        return NULL;
    }

    if (m >= n) return head;

    ListNode **backup = &head, *p = head, *q = head;

    // Move p to the mth node, q to the nth node.
    while (p != NULL && --m) backup = &p->next, p = p->next;
    while (q != NULL && --n) q = q->next;

    // If move success.
    if (m == 0 && n == 0 && p != NULL && q != NULL)
    {
        *backup = unguarded_reverse(q->next, p, q->next);
    }

    return head;
}

// Reverse every k group in linked list, return head.
// For example,
// Given this linked list: 1->2->3->4->5
// For k = 2, you should return: 2->1->4->3->5
// For k = 3, you should return: 3->2->1->4->5
ListNode *reverse_k_group(ListNode *head, int k) {
    if (head == NULL || k < 0) return NULL;

    int count = 0;

    // Count all node num.
    for (ListNode *p = head; p != NULL; p = p->next)
    {
        ++count;
    }

    for (int i = 1; i + k - 1 <= count; i += k)
    {
        head = reverse_between(head, i, i + k - 1);
    }

    return head;
}