Sharing my simple O(n) solution with explanation

  • 1
    ListNode *rotateRight(ListNode *head, int k) {
        if(head == nullptr){
            return nullptr;
        ListNode* p1 = head;
        ListNode* p2 = head;
        int size = 1;
        while(p1->next != nullptr){
            p1 = p1->next;
        p1->next = head;
        for(int i = 0; i < size - k % size - 1; i++){
            p2 = p2->next;
        p1 = p2->next;
        p2->next = nullptr;
        return p1;

    In the first For loop, I make p1 point to tail node and count the size of this linked list. Then p1->next = head make this list become a circle.

    In the second For loop, I make p2 point to the previous node before the target node. Let the previous node point to nullptr and return target node.

    Tricky place: size - k % size - 1 can make pointer point to the previous node before the target node in O(n) time.

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