25. Reverse Nodes in k-Group（K 个一组，反转链表）

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?

 

 

 

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(head == nullptr || head->next == nullptr) return head;
        ListNode* pre = nullptr;
        ListNode* cur = head;
        while(cur != nullptr) {
            ListNode* c_next = cur->next;
            cur->next = pre;
            pre = cur;
            cur = c_next;
        }
        return pre;
    }
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode* fakehead = new ListNode(-1);
        fakehead->next = head;
        ListNode* pre =fakehead;
        ListNode* cur = head;
        while(cur != nullptr) {
            for(int i = 1;i < k && cur != nullptr ;++i) {
                cur = cur->next;
            }
            if(cur==nullptr) break;
            
            ListNode* start = pre->next;
            ListNode* c_next = cur->next;

            cur->next = nullptr;
            pre->next =  reverseList(start);
            start->next = c_next;
            pre = start;
            cur = pre->next;
        }
        return  fakehead->next;
    }
};