Reverse Nodes in k-Group

## Description

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 nodes, only nodes itself may be changed.

Only constant memory is allowed. For example,

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`

## Solution

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 `````` ``````func reverseKGroup(head *ListNode, k int) *ListNode { if head == nil || k <= 1 { return head } var newHead, lastGroupTail *ListNode nextGroupHead := head for nextGroupHead != nil { right := nextGroupHead // search the first node of reversed-group for n := 0; n < k-1 && right != nil; n++ { right = right.Next } if right == nil { // no more group, so break break } if newHead == nil { // first group, set newHead newHead = right } else { // else, link last group and this group lastGroupTail.Next = right } // saving infomation to search next group lastGroupTail, nextGroupHead = nextGroupHead, right.Next // reversing this group left, right := lastGroupTail, nextGroupHead for left != nextGroupHead { left, left.Next, right = left.Next, right, left } } if newHead == nil { return head } return newHead }``````