# Reverse Nodes in k-Group

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.

```
public ListNode reverseKGroup(ListNode head, int k) {
ListNode begin;
if (head==null || head.next ==null || k==1)
return head;
ListNode dummyhead = new ListNode(-1);
dummyhead.next = head;
begin = dummyhead;
int i=0;
while (head != null){
i++;
if (i%k == 0){
begin = reverse(begin, head.next);
head = begin.next;
} else {
head = head.next;
}
}
return dummyhead.next;
}
public ListNode reverse(ListNode begin, ListNode end){
ListNode curr = begin.next;
ListNode next, first;
ListNode prev = begin;
first = curr;
while (curr!=end){
next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
begin.next = prev;
first.next = curr;
return first;
}
```

There are also recursive solutions, this video uses c++ but the approach is the same as the code below.

```
public ListNode reverseKGroup(ListNode head, int k) {
ListNode curr = head;
int count = 0;
while (curr != null && count != k) { // find the k+1 node
curr = curr.next;
count++;
}
if (count == k) { // if k+1 node is found
curr = reverseKGroup(curr, k); // reverse list with k+1 node as head
// head - head-pointer to direct part,
// curr - head-pointer to reversed part;
while (count-- > 0) { // reverse current k-group:
ListNode tmp = head.next; // tmp - next head in direct part
head.next = curr; // preappending "direct" head to the reversed list
curr = head; // move head of reversed part to a new node
head = tmp; // move "direct" head to the next node in direct part
}
head = curr;
}
return head;
}
```

## Related Problems

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists into one **sorted** list. The list should be made by splicing together the nodes of the first two lists.

Return *the head of the merged linked list*.

Given the root of a binary tree, flatten the tree into a "linked list":

The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.

The "linked list" should be in the same order as a pre-order traversal of the binary tree.

Given the root of a binary tree, return *the level order traversal of its nodes' values*. (i.e., from left to right, level by level).

A linked list of length n is given such that each node contains an additional random pointer, which could point to any node in the list, or null. Return the head of a deep copy of the list, ie all new nodes and both the node.next and node.random pointers point to new nodes.