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.

題目：倒著數k個node，從那開始到結尾和之前那部分對調，那個例子就是，4->5拿前面來，1->2->3拿後面去。

思路：我自己是用了三個指標，分別是移到k位置，第一位，然後把三段拼起來。思路是正確，但是太複雜，繞來繞去，後面提交也是超時了。

public ListNode rotateRight(ListNode head, int k) {
		
		ListNode result = new ListNode(-1);
		result.next = head;
		ListNode first = result;
		ListNode second = head;
		ListNode third = result;
		if(head == null) {
			return result.next;
		}
		if(head.next == null){
			return head;
		}
		int len = 1;
		while(result.next != null) {
			len++;
			result = result.next;
		}
		int i = 0;
		while(i < len-k) {
			i++;
			first = first.next;
		}
		third.next = first;
		int j = 1;
		while(j < len-1-k) {
			second = second.next;
			j++;
		}
		second.next = null;
		
		ListNode flag = third;
		while(flag.next != null) {
			flag = flag.next;
		}
		flag.next = head;
		
		return third.next;
		
		
	}
 

public ListNode rotateRight(ListNode head, int n) {
         if(head==null||head.next==null||n==0)
            return head;
        ListNode fast = head, slow = head,countlen = head;
        ListNode newhead = new ListNode(-1);
        int len = 0;
        
        while(countlen!=null){
            len++;
            countlen = countlen.next;
        }
        
        n = n%len;
        if(n==0)
            return head;
        
        for(int i = 0; i < n; i++)
            fast = fast.next;
        
        while(fast.next!=null){
            slow = slow.next;
            fast = fast.next;
        }
        
        newhead = slow.next;
        fast.next = head;
        slow.next = null;
        
        return newhead;
    }
 

public ListNode rotateRight(ListNode head, int n) {

    if (head == null || n == 0)
        return head;
    ListNode p = head;
    int len = 1;//since p is already point to head
    while (p.next != null) {
        len++;
        p = p.next;
    }
    p.next = head; //form a loop
    n = n % len;
    for (int i = 0; i < len - n; i++) { 
        p = p.next;
    } //now p points to the prev of the new head
    head = p.next;
    p.next = null;
    return head;
}

種一棵樹最好的時間是十年前，其次是現在！