链表反转的几种形式

// 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->5ListNode *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;}