思路：
模拟。
实现：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode() : val(0), next(nullptr) {} 7 * ListNode(int x) : val(x), next(nullptr) {} 8 * ListNode(int x, ListNode *next) : val(x), next(next) {} 9 * }; 10 */ 11 class Solution 12 { 13 public: 14 ListNode* reverse(ListNode* head) 15 { 16 if (head == NULL or head->next == NULL) return head; 17 ListNode* pre = head, *cur = head->next; 18 head->next = NULL; 19 while (cur) 20 { 21 ListNode* nxt = cur->next; 22 cur->next = pre; 23 pre = cur; 24 cur = nxt; 25 } 26 return pre; 27 } 28 ListNode* reverseKGroup(ListNode* head, int k) 29 { 30 ListNode* cur = head, *cur_head = head, *res = NULL, *last_tail = NULL; 31 int cnt = 1; 32 while (cur) 33 { 34 if (cnt % k == 0) 35 { 36 ListNode* next_head = cur->next; 37 cur->next = NULL; 38 ListNode* new_head = reverse(cur_head); 39 if (res == NULL) res = new_head; 40 if (last_tail) last_tail->next = new_head; 41 cur_head->next = next_head; 42 last_tail = cur = cur_head; 43 cur_head = cur->next; 44 } 45 cnt++; 46 cur = cur->next; 47 } 48 return res ? res: head; 49 } 50 };