In this Leetcode Rotate List problem solution we have given the head of a linked list, rotate the list to the right by k places.

leetcode rotate list problem solution


Problem solution in Python.

class Solution:
    def rotateRight(self, head: ListNode, k: int) -> ListNode:
        if head == None: return 
        values = []
        dummay = ListNode()
        cur = dummay
        while head:
            values.append(head.val)
            head = head.next
        for i in range(k % len(values)):
            values.insert(0,values.pop())
        for j in values:
            cur.next = ListNode(j)
            cur = cur.next
        return dummay.next



Problem solution in Java.

public class Solution {
public ListNode rotateRight(ListNode head, int k) {
    if(head == null || head.next == null) return head;
    ListNode p = head;
    int count = 1;
    while(p.next!= null){
        count++;
        p = p.next;
    }
    k = k % count;
    p.next = head;      /*make a circle here*/
    p = head;
    for(int i = 0; i < count - k - 1; i++){
        p = p.next;
    }
    ListNode dummy = p.next;
    p.next = null;
    return dummy;
}
}


Problem solution in C++.

class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if(head == NULL || head->next == NULL) return head;
        int nodes = 1;
        ListNode *temp = head;
        while(temp->next != NULL){
            temp = temp->next;
            nodes++;
        }
        k = k % nodes;
        if(k == 0)return head;
        int i = 0;
        ListNode *cur = head;
        while(i<nodes-k-1){
            cur = cur->next;
            i++;
        }
        ListNode *newHead = cur->next;
        temp->next = head;
        cur->next = NULL;
        return newHead;
    }
};


Problem solution in C.

struct ListNode* rotateRight(struct ListNode* head, int k){
    int length = 0;
    struct ListNode* tmp;
    struct ListNode* tmp2;
    tmp = head;
    while (tmp) {
        length++;
        tmp2 = tmp;
        tmp = tmp->next;
    }
    if (length == 0) {
        return head;
    }
    k = k % length;
    tmp = head;
    for (int i = 0; i < length - k - 1; i++) {
        tmp = tmp->next;
    }
    tmp2->next = head;
    head = tmp->next;
    tmp->next = NULL;
    return head;
}