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 <= 50000 <= Node.val <= 10001 <= k <= sz
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 };