25. Reverse Nodes in k-Group

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.

You may not alter the values in the list's nodes, only nodes themselves may be changed.

 

Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]

Example 2:

Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]

Example 3:

Input: head = [1,2,3,4,5], k = 1
Output: [1,2,3,4,5]

Example 4:

Input: head = [1], k = 1
Output: [1]

 

Constraints:

• The number of nodes in the list is in the range sz.
• 1 <= sz <= 5000
• 0 <= Node.val <= 1000
• 1 <= k <= sz

 

Follow-up: Can you solve the problem in O(1) extra memory space?  

 

 

   

 1 class Solution {
 2 public:
 3     ListNode* reverseList(ListNode* head) {
 4         if(head == nullptr || head->next == nullptr) return head;
 5         ListNode* pre = nullptr;
 6         ListNode* cur = head;
 7         while(cur != nullptr) {
 8             ListNode* c_next = cur->next;
 9             cur->next = pre;
10             pre = cur;
11             cur = c_next;
12         }
13         return pre;
14     }
15     ListNode* reverseKGroup(ListNode* head, int k) {
16         ListNode* fakehead = new ListNode(-1);
17         fakehead->next = head;
18         ListNode* pre =fakehead;
19         ListNode* cur = head;
20         while(cur != nullptr) {
21             for(int i = 1;i < k && cur != nullptr ;++i) {
22                 cur = cur->next;
23             }
24             if(cur==nullptr) break;
25             
26             ListNode* start = pre->next;
27             ListNode* c_next = cur->next;
28 
29             cur->next = nullptr;
30             pre->next =  reverseList(start);
31             start->next = c_next;
32             pre = start;
33             cur = pre->next;
34         }
35         return  fakehead->next;
36     }
37 };