CODEWITHSUNDEEP
algorithmlinked-listspace-complexity
Poulami
Poulami

Reputation: 1087

Reversing a singly linked list when a block size is given

There is a singly connected linked list and a block size is given.For eg if my linked list is 1->2->3->4->5->6->7->8-NULL and my block size is 4 then reverse the first 4 elements and then the second 4 elements.The output of the problem should be 4->3->2->1->8->7->6->5-NULL

I was thinking of dividing the linked list into segments of size 4 and then reversing it. But that way I am forced to use a lot of extra nodes which is not desired at all. The space complexity should be kept to a minimum.

It will be highly appreciable if someone can come with a better solution where the usage of extra nodes would be kept to a minimum.

Upvotes: 4

Views: 1633

Answers (3)

vine'th
vine'th

Reputation: 5030

This can be done in linear-time, with constant space. Here is a brief description:

  1. Split the linked list into two parts by block-size

    
int split(node* head, node** listA, node** listB size_t block_size)
{
node* cur = head;

while(block_size && cur)
{
   cur = cur->next;
   --block_size;
}
if(!cur) { /* error : invalid block size */ return -1; }
*listA = head;
*listB = cur->next;
cur->next = NULL; /* terminate list A */
return 0;
}

  2. Reverse the two sub-parts, (use a non-recursive linear time, constant space function)

    
reverse(listA);
reverse(listB);

  3. Link them to get the desired linked list.

    
cur = *listA;
/* goto last but one element of listA */
while(cur->next) cur = cur->next; 
cur->next = *listB;

Upvotes: 0

josh
josh

Reputation: 14463

I tried this...seems to work fine...

node* reverse(node* head) // function to reverse a list
{
    node* new_head = NULL;
    while(head != NULL)
    {
        node* next = head->next;
        head->next = new_head;
        new_head = head;
        head = next;
    }
    return new_head;
}

node* reverse_by_block(node* head, int block)
{
    if(head == NULL)
            return head;

    node* tmp = head;
    node* new_head = head;
    node* new_tail = NULL;

    int count = block;
    while(tmp != NULL && count--)
    {
        new_tail = tmp;
        tmp = tmp->next;
    }

    new_tail->next = NULL;
    new_tail = new_head;
    new_head = reverse(new_head);
    new_tail->next = reverse_by_block(tmp,block);

    return new_head;
}

Upvotes: 2

LaC
LaC

Reputation: 12824

You can advance swapping the current element with the next 3 times: 1234, 2134, 2314, 2341. Then do it twice to get 3421. Then once to get 4321. Then advance 4 steps and repeat the process with the next block.

Upvotes: 0

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