61. Rotate List 再做一次

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.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if(head==null||head.next==null) return head;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode slow = dummy; ListNode fast = dummy;
        
        int i;
        for(i=0;fast.next!=null;i++){
            fast = fast.next;
        }
        
        //move to the node that start to need to be moved
        for(int j=i-k%i;j>0;j--){
            slow = slow.next;//循環移動後的第一個節點，它的下一個節點作爲第一個節點
        }
        
        fast.next=dummy.next; //Do the rotation
        dummy.next=slow.next; 
        slow.next=null;
        
        return dummy.next;
    }
}

這裏的重點是：

1. 判斷條件 是 len-k%len