leetcode 旋转链表 mid

题目链接

链表不能通过index来访问，要一步步的走，所以要麻烦一些。

要注意当k大于链表长度时的处理，需要先遍历一遍得到链表长度n，然后k对n取余，用余数来翻转。

/**
 * 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) return null;
        ListNode Head = new ListNode(0);
        Head.next = head;
        ListNode f = Head, s = Head;
        int len = 0;
        while(f.next != null) {
            len++;
            f = f.next;
        }
        //System.out.println(Head.next.val);
        k %= len;
        for(int j = len - k; j > 0; j--) 
            s = s.next;
        f.next = Head.next;
        //System.out.println(Head.next.val);
        Head.next = s.next;
        //System.out.println(s.next.val);
        s.next = null;
        return Head.next;
    }
}