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# Reverse Nodes in k-Group LeetCode Solution

## Problem – 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`

Follow-up: Can you solve the problem in `O(1)` extra memory space?

## Reverse Nodes in k-Group LeetCode Solution in Java

``````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
}
}
}
``````

## Reverse Nodes in k-Group LeetCode Solution in C++

``````class Solution
{
public:

ListNode* reverse(ListNode* first, ListNode* last)
{
ListNode* prev = last;

while ( first != last )
{
auto tmp = first->next;
first->next = prev;
prev = first;
first = tmp;
}

return prev;
}

ListNode* reverseKGroup(ListNode* head, int k)
{
for (int i=0; i < k; ++i)
{
if ( ! node  )
return head; // nothing to do list too sort
node = node->next;
}

head->next = reverseKGroup( node, k);
}
};
``````

## Reverse Nodes in k-Group LeetCode Solution in Python

``````def reverseKGroup(self, head, k):
dummy = jump = ListNode(0)
dummy.next = l = r = head

while True:
count = 0
while r and count < k:   # use r to locate the range
r = r.next
count += 1
if count == k:  # if size k satisfied, reverse the inner linked list
pre, cur = r, l
for _ in range(k):
cur.next, cur, pre = pre, cur.next, cur  # standard reversing
jump.next, jump, l = pre, l, r  # connect two k-groups
else:
return dummy.next
``````
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