Rotate List - LeetCode

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

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.

没有完全AC，还有细节小问题，但是大体逻辑是这样的。

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode rotateRight(ListNode head, int n) {       if(head == null){            return null;        } if(n == 0)        return head;        ListNode newhead = new ListNode(0);        int count = 1;        ListNode current = new ListNode(0);        current = head;                while(current.next != null){            count ++;            current = current.next;        }      //  System.out.println(count);        if(count == 1)        return head;    if(n > count)    n = count;        ListNode tail = current;        current = head;                for(int i = 1; i < count - n; i++){            current = current.next;        }        newhead.next = current.next;        tail.next = head;        current.next = null;          return newhead.next;    }    }

参考的AC代码：

public class Solution {    private int getLength(ListNode head) {        int length = 0;        while (head != null) {            length ++;            head = head.next;        }        return length;    }        public ListNode rotateRight(ListNode head, int n) {        if (head == null) {            return null;        }                int length = getLength(head);        n = n % length;                ListNode dummy = new ListNode(0);        dummy.next = head;        head = dummy;                ListNode tail = dummy;        for (int i = 0; i < n; i++) {            head = head.next;        }                while (head.next != null) {            tail = tail.next;            head = head.next;        }                head.next = dummy.next;        dummy.next = tail.next;        tail.next = null;        return dummy.next;    }}