61. Rotate List

Description

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

Problem URL

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Solution

重复k次，将链表最后一个节点拿到头上。

Using first list node to get the length of the linked list. And it is at the tail node, then calculate the result tail position by length - k % length. Using second list node to get there. set the rest node to head and set second.next to null.

Code

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if (head == null || head.next == null || k == 0){
            return head;
        }
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode first = dummy;
        ListNode second = dummy;
        
        int length = 0;
        while (first.next != null){
            length++;
            first = first.next;
        }
        
        int count = length - k % length;
        while(count-- != 0){
            second = second.next;
        }
        
        first.next = dummy.next;
        dummy.next = second.next;
        second.next = null;
        return dummy.next;
        
    }
}

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

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