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.


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.


The logic in the below code involves breaking up the list into k parts and reversing them individually.

func reverseKGroup(head *ListNode, k int) *ListNode {
    th := &ListNode{Val : 0, Next: head}
    cur := th
    if (k<2) {return head}
    for cur.Next != nil{
        start := cur.Next
        end := cur.Next
        for i:=1;i<k&&end!=nil;i++{ end = end.Next } // move end k times
        if end==nil{
        next := end.Next
        end.Next = nil
        // reverse the kth part of the list
        t:= start
        var prev *ListNode
        for t!=nil{
            next := t.Next
            t.Next = prev
            prev = t
            t = next
        cur.Next = end
        start.Next = next
        cur = start
    return th.Next