Rotating linked list (illustration)

Leetcode Title Description ：
Topic link ： Rotate the list

The title is very concise , It's the mobile node , But there is an implied meaning , To move each node of the linked list to the right , Make each node in a new location , Because there is no node behind the tail node , To move, you need to form a ring , All nodes turn in the ring , After moving, disconnect the ring , Form a new linked list . Method ： Circular list + Move

The basic idea ：

1. Calculate the length of the list len, Because when calculating the length, we found the last node of the linked list , The length of the last node does not count , therefore len Starting at 1.
   2. Observe the operation of the linked list every len One cycle at a time , Calculation k % len It can effectively reduce the number of times to operate the linked list ( If the chain length is 3, You move each node of the linked list 4 A place , In fact, it just moves each node 1 A place )
2. Find the end node of the linked list , Modify it next The pointer , Connect the linked list into a circular linked list , When calculating the length, you can turn the linked list into a circular linked list at the same time .
3. Loop through the linked list from the tail node , Until you find the tail node of the linked list after rotation , Put it next Domain set to null, Disconnect the circular linked list from here And return the new head node , Traverse from the tail node len-k%len Once, you can find the tail node of the new linked list , Then find the head node .

Graphic analysis ：

We can deduce through one example or several example classes , Conditions for finding the head node precursor after moving the node .

Dynamic diagram demonstration ：

The above question is my notes and summary , There are various ways to solve problems with linked lists , This is just a basic method . It's also very simple , It's not as hard to think about using two finger acupuncture , This method is only for the reference of Xiaobai like me .
Reference code ：

public Node rotateRight(Node  head, int k) {
    
        if (head == null || head.next == null) {
    
            return head;
        }
        int len=1;
        int index;
        Node  temp = head;
        Node newHead;
        while (temp.next != null) {
     // Calculate the length of the list len, The condition here is that the last node is found , The last node is not calculated , therefore len from 1 Start 
            len++;
            temp = temp.next;
        }
        /*  Usually we find the length , Is to traverse the entire linked list , The condition for finding the length should be temp!=null, But for this problem, our code above not only finds the length, but also finds the last node of the linked list ,  Make the last node next  The field points to the head node and forms a ring , Of course, you can also calculate the length of the linked list normally , But after that, it will take an extra procedure to find the tail node of the linked list . int len=0; while (temp!=null) { len++; temp=temp.next; }*/
        temp.next = head; // Connect the linked list into a circular linked list 
        k %= len; // Rotate the linked list every len One cycle at a time , So calculate k Yes len Surplus of , Avoid repetition 
        index = len - k; //
        while (index-- > 0) {
    
            temp = temp.next;// Find the node to disconnect the circular linked list ,
        }
        newHead = temp.next;// Record the new head node of the linked list 
        temp.next = null; // Disconnect the circular linked list 
        return newHead;
    }

️ Too small 🧑*‍‍🧑*, Remember to hold ,