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Reverse Nodes in k-Group | Leetcode 25 solution

 25. Reverse Nodes in k-Group | Leetcode Solution



Given the head of a linked list, reverse the nodes of the list k at a time, and return the 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]

 

Constraints:

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


Solution :




/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode() : val(0), next(nullptr) {}
*     ListNode(int x) : val(x), next(nullptr) {}
*     ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
 
    ListNode* solve(ListNode* head, int k , int cnt)
    { 
        if(cnt < k )
        {
            return head;
        }
        if(head  == NULL )
        {
            return head;
        }
        int t =0;
        ListNode* curr = head;
        ListNode* pre = NULL ;
        ListNode* nex = NULL;
        ListNode* x =NULL;
        while(cnt >= k && t < k )
        { 
            if(t == 0)
            {
                x = curr;
            }
            nex = curr ->next;
            curr ->next = pre;
            pre = curr;
            curr = nex;
            t++;
        }
        x ->next =  solve(curr  , k, cnt - k);
        return pre;

    }
    ListNode* reverseKGroup(ListNode* head, int k) {

        ListNode* temp = head;
        int cnt = 0;
        while(temp != NULL)
        {
            cnt++;
            temp = temp->next;
        }
        return solve(head , k , cnt);
 

    }
};



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