k个一组反转链表

• 思路：k_reverse取k个节点反转，剩下的再调用k_reverse取k个节点反转，直到节点数不等于k
• 关键在于理解part_reverse后的k个节点是end开头，head结尾，所以前k个节点的head指向后k个节点的end（已经反转的）
• 想不清楚可以先看循环版本

struct link
{
    int a;
    link* next;
};

link* part_reverse(link *begin, link *end) {
    link *pre = end->next;
    while(pre != end) {
        link *tmp = begin->next;
        begin->next = pre;
        pre = begin;
        begin = tmp;
    }
    return pre;
}

link* k_reverse(link* head, int k) {
    int cnt = 0;
    link * end = head;
    while(end != NULL) {
        ++cnt;
        if(cnt == k) break;
        end = end->next;
    }
    if(cnt != k) return head;
    link *tmp = part_reverse(head, end);
    head->next = k_reverse(head->next, k);
    return tmp;
}

link* part_reverse(link *begin, link *end) {
    link *pre = end->next;
    while(pre != end) {
        link *tmp = begin->next;
        begin->next = pre;
        pre = begin;
        begin = tmp;
    }
    return pre;
}

link* k_reverse(link* head, int k) {
    int cnt = 0;
    link *end = head;
    while (end != NULL)
    {
        ++cnt;
        if(cnt == k) break;
        end = end->next;
    }
    if(cnt != k) return head;
    link *res = part_reverse(head, end);
    while(1) {
        cnt = 0;
        link *pre = head;
        head = pre->next;
        end = head;
        while(end != NULL) {
            ++cnt;
            if(cnt == k) break;
            end = end->next;
        }
        if(cnt != k) return res;
        pre->next = part_reverse(head, end);
    }
}