LeetCode 61 - Rotate a Linked List | GeeksforGeeks

http://n00tc0d3r.blogspot.com/2013/05/rotate-list.html
Rotate a Linked List Given a singly linked list, rotate the linked list counter-clockwise by k nodes.
For example:
Given 1->2->3->4->5->NULL and k = 2, return 4->5->1->2->3->NULL.
Given 1->2->3->4->5->NULL and k = 6, return 5->1->2->3->4->NULL.
aaa

There are two cases: k < length-of-list and k >= length-of-list.
In the case of k < length-of-list,

• use two pointers to find the k-to-the-end node,
• link the original tail to the original head,
• then cut the list before k-to-the-end node and make it the new head.

In the other case, again, we use two pointers to iterate through the list. When we hit the tail for the first time, link it to the original head to make the list into a circular linked list and then keep forwarding the pilot pointer until we find the k-to-the-end node.
Note that now the tail node is the one whose next is head (not null any more).

public ListNode rotateRight(ListNode head, int k) {  
   if (head == null || k <= 0 || head.next == null) return head;  
   
   ListNode pre = head, end = head;  
   // find the k-to-the-end  
   while (end.next != head || k > 0) {  
     if (k > 0) { // only forward end pointer  
       --k;  
     } else { // forward both pointers  
       pre = pre.next;  
     }  
     end = end.next;  
     // make it a circular linked list  
     if (end.next == null) end.next = head;  
   }  
   
   ListNode newHead = pre.next;  
   pre.next = null;  
   return newHead;  
 }  




In this algorithm, the pilot pointer has been forwarded at least k times and the two pointers were moved together to get to the k-to-the-end node. The worst case running time is O(k+n) where n is the length of the list.

Ask the interviewer, or consider different possible input cases: k >> n.
If k >> n, we can improve the algorithm by first compute the length of the list. By doing this, the running time becomes O(2n) = O(n) since we iterate through the list twice: one for the length and the other for the k-to-the-end node.

https://leetcode.com/problems/rotate-list/discuss/22715/Share-my-java-solution-with-explanation

public ListNode rotateRight(ListNode head, int n) {
    if (head==null||head.next==null) return head;
    ListNode dummy=new ListNode(0);
    dummy.next=head;
    ListNode fast=dummy,slow=dummy;

    int i;
    for (i=0;fast.next!=null;i++)//Get the total length 
     fast=fast.next;
    
    for (int j=i-n%i;j>0;j--) //Get the i-n%i th node
     slow=slow.next;
    
    fast.next=dummy.next; //Do the rotation
    dummy.next=slow.next;
    slow.next=null;
    
    return dummy.next;
}

http://rleetcode.blogspot.com/2014/01/rotate-list-java.html

public ListNode rotateRight(ListNode head, int k) {  
   // find the length of the list  
   int len = 0; ListNode cur = head;  
   while (cur != null) {  
     ++len;  
     cur = cur.next;  
   }  
   
   if (len == 0 || k % len == 0) return head;  
   k = k % len;  
   
   ListNode pre = head; cur = head;  
   // find the n-to-the-end  
   while (cur.next != head) {  
     if (k > 0) { // only forward end pointer  
       --k;  
     } else { // forward both pointers  
       pre = pre.next;  
     }  
     cur = cur.next;  
     // make it a circular linked list  
     if (cur.next == null) cur.next = head;  
   }  
   
   ListNode newHead = pre.next;  
   pre.next = null;  
   
   return newHead;  
 } 

void rotate (struct node **head_ref, int k)

{

     if (k == 0)

       return;

    // Let us understand the below code for example k = 4 and

    // list = 10->20->30->40->50->60.

    struct node* current = *head_ref;

    // current will either point to kth or NULL after this loop.

    //  current will point to node 40 in the above example

    int count = 1;

    while (count < k && current != NULL)

    {

        current = current->next;

        count++;

    }

    // If current is NULL, k is greater than or equal to count

    // of nodes in linked list. Don't change the list in this case

    if (current == NULL)

        return;

    // current points to kth node. Store it in a variable.

    // kthNode points to node 40 in the above example

    struct node *kthNode = current;

    // current will point to last node after this loop

    // current will point to node 60 in the above example

    while (current->next != NULL)

        current = current->next;

    // Change next of last node to previous head

    // Next of 60 is now changed to node 10

    current->next = *head_ref;

    // Change head to (k+1)th node

    // head is now changed to node 50

    *head_ref = kthNode->next;

    // change next of kth node to NULL

    // next of 40 is now NULL

    kthNode->next = NULL;

}

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