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Rotate List LeetCode Solution

Problem – Rotate List

Given the head of a linked list, rotate the list to the right by k places.

Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [4,5,1,2,3]

Example 2:

Input: head = [0,1,2], k = 4
Output: [2,0,1]


  • The number of nodes in the list is in the range [0, 500].
  • -100 <= Node.val <= 100
  • 0 <= k <= 2 * 109

Rotate List LeetCode Solution in C++

class Solution {
    ListNode* rotateRight(ListNode* head, int k) {
        if(!head) return head;
        int len=1; // number of nodes
        ListNode *newH, *tail;
        while(tail->next)  // get the number of nodes in the list
            tail = tail->next;
        tail->next = head; // circle the link

        if(k %= len) 
            for(auto i=0; i<len-k; i++) tail = tail->next; // the tail node is the (len-k)-th node (1st node is head)
        newH = tail->next; 
        tail->next = NULL;
        return newH;

Rotate List LeetCode Solution in Java

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

    int i;
    for (i=0;!=null;i++)//Get the total length;
    for (int j=i-n%i;j>0;j--) //Get the i-n%i th node;; //Do the rotation;;

Rotate List LeetCode Solution in Python

class Solution:
    def rotateRight(self, head: ListNode, k: int) -> ListNode:
        if not head:
            return None
        lastElement = head
        length = 1
        # get the length of the list and the last node in the list
        while ( ):
            lastElement =
            length += 1

        # If k is equal to the length of the list then k == 0
        # ElIf k is greater than the length of the list then k = k % length
        k = k % length
        # Set the last node to point to head node
        # The list is now a circular linked list with last node pointing to first node = head
        # Traverse the list to get to the node just before the ( length - k )th node.
        # Example: In 1->2->3->4->5, and k = 2
        #          we need to get to the Node(3)
        tempNode = head
        for _ in range( length - k - 1 ):
            tempNode =
        # Get the next node from the tempNode and then set the as None
        # Example: In 1->2->3->4->5, and k = 2
        #          tempNode = Node(3)
        #          answer = Node(3).next => Node(4)
        #          Node(3).next = None ( cut the linked list from here )
        answer = = None
        return answer
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