[leetcode] Rotate List

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

For example:
Given1->2->3->4->5->NULLand k =2,
return4->5->1->2->3->NULL.

https://oj.leetcode.com/problems/rotate-list/

思路1：先遍历一遍求长度，然后从头再开始走k%len步，设置新的头尾节点，连接原来的头尾节点。

思路2：遍历求长度，将首尾连接，继续走到新的头尾，断开。

思路1代码：

public class Solution {
    public ListNode rotateRight(ListNode head, int n) {
        if (head == null)
            return null;
        int len = 0;
        ListNode p = head;
        while (p != null) {
            len++;
            p = p.next;
        }
        
        p = head;
        int i;
        n = n % len;
        if (n == 0)
            return head;
        for (i = 0; i < len - n - 1; i++)
            p = p.next;
        ListNode newTail = p;
        ListNode newHead = p.next;

        while (p.next != null)
            p = p.next;
        p.next = head;
        newTail.next = null;
        return newHead;

    }

    public static void main(String[] args) {
        ListNode head = ListUtils.makeList(1, 2, 3, 4, 5);
        ListUtils.printList(head);
        head = new Solution().rotateRight(head, 2);
        ListUtils.printList(head);
    }
}

第二遍记录：

　　改了个小地方，第一遍求长度的时候，顺便把头尾连接起来。

public class Solution {
    public ListNode rotateRight(ListNode head, int n) {
        if(head==null)
            return null;
        int len=1;
        ListNode p = head;
        while(p.next!=null){
            len++;
            p=p.next;
        }
        p.next=head;
        
        n = n%len;
        p=head;
        for(int i=0;i<len-1-n;i++){
            p=p.next;
        }
        ListNode newHead =p.next;
        p.next=null;
        return newHead;
    }
}