题解 | #链表中的节点每k个一组翻转#

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 *  ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
#include <cstddef>
#include <ostream>
class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param head ListNode类
     * @param k int整型
     * @return ListNode类
     */
    ListNode* reverseKGroup(ListNode* head, int k) {
        if (head == NULL) {
            return NULL;
        }
        ListNode* num = new ListNode(-1);
        num = head;
        int j = 0;
        while (num != NULL) {
            num = num->next;
            j++;
        }
        int cut = j / k;

        ListNode* res = new ListNode(-1);
        res->next = head;
        ListNode* end = head;
        ListNode* start = res;
        for (int z = 0; z < cut; z++) {
            for (int i = 1; i < k; i++) {
                ListNode* temp = end->next;
                end->next = temp->next;
                temp->next = start->next;
                start->next = temp;
            }
            start=end;
            end=end->next;
        }
        return res->next;

    }
};