题解 | #链表中的节点每k个一组翻转#

先求出链表长度，得到需要翻转的次数（len/k）
调用翻转函数，进行翻转；

class Solution {
public:
    /**
     *
     * @param head ListNode类
     * @param k int整型
     * @return ListNode类
     */
    ListNode* revser(ListNode* head, int n) {
    ListNode* newhead = new ListNode(-1);
    ListNode* flag = head;
    while (head&&n > 0) {
        ListNode* temp = head->next;
        head->next = newhead->next;
        newhead->next = head;
        head = temp;
        n--;
    }
    flag->next = head;
    return newhead->next;
}

ListNode* reverseKGroup(ListNode* head, int k) {
    // write code here
    ListNode* newhead = new ListNode(-1);
    newhead->next = head;
    ListNode* first = head;
    int len = 0;
    while (first) {
        first = first->next;
        len++;
    }
    first = newhead;
    for (int i = 0; i < len / k; i++) {
        ListNode* pre = first->next;
        first->next = revser(first->next, k);
        first = pre;
    }
    return newhead->next;
}
};