Reverse Nodes in k-Group

Jin Shang bio photo By Jin Shang

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.


Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5


  • Only constant extra memory is allowed.
  • You may not alter the values in the list’s nodes, only nodes itself may be changed.


Use recursion: (head is the current recursion node, cur is supposedly next)

1) if head == NULL, return

2) if depth == k, then prev->next is head, save head->next to next, found = true, save head to cur;

3) else do recursion: if(found) then cur->next = head.

4) if depth == 1: if(found) head->next is saved next, reset parameters, then do recursion; else return.

5) save head to cur

 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
class Solution {
    void rK(ListNode* head, int depth, ListNode* prev, ListNode*& next, int k, bool &found, ListNode*& cur){
        if(head == NULL){
        if(depth == k){
            prev->next = head; 
            next = head->next;
            found = true;
            cur = head;
        rK(head->next, depth+1, prev, next, k, found, cur);
            cur->next = head;
                head->next = next;
                next = NULL;
                found = false;
                cur = NULL;
                rK(head->next, 1, head, next, k, found, cur);
        cur = head;
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* next = NULL;
        bool found = false;
        ListNode* cur = NULL;
        rK(head, 1, dummy, next, k, found, cur);
        return dummy->next;


#Easy reverse
end = last
while(head != last){
	temp = head->next;
    head->next = end;
    end = head;
    head = temp;