Leetcode Reverse Nodes in k-Group problem solution

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:
        dummy_head=M_head=ListNode()
        dummy_head.next=head
       # Calculate Length
        l=0
        t=head
        
        while t:
            l+=1
            t=t.next
            
        # Get multiple length of k
        l//=k
        
        while head and l:
            
            for _ in range(1,k):
                
                temp=head.next
                head.next=temp.next
                temp.next=M_head.next
                M_head.next=temp
                
            if head:
                M_head,head=head,head.next
            l-=1
        return dummy_head.next

Problem solution in Java.

public ListNode reverseKGroup(ListNode head, int k) {
        if(head == null) return head;
        ListNode tail = head;
        int movements = k;
        while(tail != null && movements > 1){
            tail = tail.next;
            movements--;
        }
        if(tail == null) return head;
        
        ListNode next = tail.next;
        tail.next = null;
        ListNode newHead = reverse(head);
        head.next = reverseKGroup(next, k);
        return newHead;
    }
    public ListNode reverse(ListNode head){
        if(head == null || head.next == null) return head;
        ListNode prev = null;
        ListNode cur = head;
        while(cur != null){
            ListNode nextTemp = cur.next;
            cur.next = prev;
            prev = cur;
            cur = nextTemp;
        }
        return prev;
    }

Problem solution in C++.

class Solution {
public:
    
    ListNode* reverse(ListNode* head) {
        if(head==NULL || head->next==NULL)
            return head;
        ListNode* temp=reverse(head->next);
        ListNode* newHead=temp;
        while(temp->next!=NULL)temp=temp->next;
        temp->next=head;
        head->next=NULL;
        return newHead;
    }
    ListNode* reverseKGroup(ListNode* head, int k) {
        if(k==0 || head==NULL || head->next==NULL)
            return head;
        int cnt=k-1;
        ListNode* tmp=head;
        while(cnt && tmp->next!=NULL){
            tmp=tmp->next;cnt--;
        }
        if(cnt!=0)
            return head;
         ListNode* t=reverseKGroup(tmp->next,k);
        tmp->next=NULL;
         ListNode* newHead=reverse(head);
        tmp=newHead;
        while(tmp->next!=NULL)
            tmp=tmp->next;
        tmp->next=t;
        return newHead;
        
        
    }
};

Problem solution in C.

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

    if (head) return head;
    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;
        head = prev;
        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;
            }
        }
    }
    return head;
}

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