In this Leetcode Reverse Nodes in k-Group problem solution, we have 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 list's nodes, only nodes themselves may be changed.

Problem solution in Python.

class Solution:
def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
# Calculate Length
l=0

while t:
l+=1
t=t.next

# Get multiple length of k
l//=k

for _ in range(1,k):

l-=1

Problem solution in Java.

public ListNode reverseKGroup(ListNode head, int k) {
int movements = k;
while(tail != null && movements > 1){
tail = tail.next;
movements--;
}

ListNode next = tail.next;
tail.next = null;
}
ListNode prev = null;
while(cur != null){
ListNode nextTemp = cur.next;
cur.next = prev;
prev = cur;
cur = nextTemp;
}
return prev;
}

Problem solution in C++.

class Solution {
public:

while(temp->next!=NULL)temp=temp->next;
}
ListNode* reverseKGroup(ListNode* head, int k) {
int cnt=k-1;
while(cnt && tmp->next!=NULL){
tmp=tmp->next;cnt--;
}
if(cnt!=0)
ListNode* t=reverseKGroup(tmp->next,k);
tmp->next=NULL;
while(tmp->next!=NULL)
tmp=tmp->next;
tmp->next=t;

}
};

Problem solution in C.

struct ListNode* kSteps(struct ListNode* head, int k) {
for (int i = 1; i < k && head; i++)

return NULL;
}

struct ListNode* reverseKGroup(struct ListNode* head, int k) {
if (k < 2) return head;
struct ListNode* prev = kSteps(head, k);
if (prev) {
struct ListNode* next = head, * curr = head, * tail = curr;
prev = prev->next;
for (int i = 0; i < k; i++) {
next = next->next;
curr->next = prev;
prev = curr;
curr = next;
}
while (prev = kSteps(curr, k)) {
tail->next = prev;
tail = curr;
prev = prev->next;
for (int i = 0; i < k; i++) {
next = next->next;
curr->next = prev;
prev = curr;
curr = next;
}
}
}