题解 | #链表中的节点每k个一组翻转#

/**
 * struct ListNode {
 *    int val;
 *    struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     *
     * @param head ListNode类
     * @param k int整型
     * @return ListNode类
     */
     ListNode* reverseKGroup(ListNode* head, int k) {
         ListNode* node = new ListNode(-1);
         node->next = head;
         int length = 0;
         while(head) {
             head = head->next;
             length++;
         }
         int n = length / k; // 翻转的子数组数

         auto tail = node;
         auto cur = node->next;
         while(n--) { // 一共执行n次子数组遍历
             int count = k - 1; // 遍历到要翻转的链表元素的前一个元素
             while(cur && count--) {
                 auto pre = cur->next; // pre为要翻转的最后一个元素
                 cur->next = pre->next; // 将pre头插到tail节点后
                 pre->next = tail->next;
                 tail->next = pre;
             }
             tail = cur;
             cur = cur->next;
         }
         return node->next;
     }
};