k个一组反转链表

一、递归

• 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）

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;
}

二、循环