Reverse Nodes in k-Group

Reverse Nodes in k-Group

Given a linked list, reverse the nodes of a linked list k at a time and return its modified 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

http://collabedit.com/axmt2

/*
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
*/

Node* rev (Node* head, int k)
{
    if(k<=1) return;
    
    Node* new_head = head;
    bool head_flag = false;
    
    while(true)
    {
        if(head==NULL) return new_head ;
        if(head->next==NULL) return new_head ;
        
        // Find the k-th node
        Node* tail = head;
        for(int i=0; i<k-1; i++)
        {
            if(tail->next==NULL) return new_head;
            tail=tail->next;
        } 
        if(!head_flag) {head_flag = true; new_head = tail;}
        
        // Reverse these k nodes
        Node* h = head;
        Node* c = h->next, p = head, n;
        
        while(c!=tail)
        {
            n = c->next;
            c->next = p;
            p = c;
            c = n;
        }    
        h->next = tail->next;
        tail->next = p;          
        
        // Prepare for the next iteration
        head = h->next; 
    }    
}