[LeetCode October Challange] Day 7 - Rotate List

Given a linked list, rotate the list to the right by k places, where k is non-negative.

Example 1:

Input: 1->2->3->4->5->NULL, k = 2
Output: 4->5->1->2->3->NULL
Explanation:
rotate 1 steps to the right: 5->1->2->3->4->NULL
rotate 2 steps to the right: 4->5->1->2->3->NULL

Example 2:

Input: 0->1->2->NULL, k = 4
Output: 2->0->1->NULL
Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right: 0->1->2->NULL
rotate 4 steps to the right: 2->0->1->NULL

---

Solution

Time complexity : O(n)
Space complexity : O(1)

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if (!head) return nullptr;
        else if (!head->next) return head;
        
        // find connect point & count len
        ListNode* old_tail = head;
        int len = 1;
        for (;old_tail->next; old_tail=old_tail->next)
            ++len;
        old_tail->next = head;
        
        // find new tail, avoid dulplicate operation by mod operator.
        ListNode* new_tail = head;
        for (int i=1; i<len-k%len; ++i)
            new_tail = new_tail->next;
        head = new_tail->next;

        new_tail->next = nullptr;
        return head;
    }
};