Medium
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.
终于又不看答案写出来一道medium, 渣渣就是这么容易满足。这道题一开始也美看懂那个k指的是什么，后来用了几个Custom Testcase自己看了下，意思是从链表尾部往前数k个节点一起按原顺序移动到链表首部。
首先通过curt遍历链表来计算节点总数total, 当k > total的时候实际上移动了k % total个元素，所以干脆让k = k % total. 同时prev被移动到了原链表尾部。根据题意，原链表尾部肯定会接上原链表首部，所以让prev.next = head.
然后我们来找rotate后链表的头。用prev来遍历，count来计数，当count == total - k的时候，prev.next就是新链表头所在的位置。然后我们让之前保存的dummy的next等于prev.next, 就设置好了新的链表头，返回dummy.next即可。
WechatIMG53.jpeg
/**
* 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 || head.next == null || k == 0){
return head;
}
ListNode dummy = new ListNode(-1);
dummy.next = head;
ListNode prev = dummy;
ListNode curt = head;
int total = 0;
while (curt != null){
total++;
curt = curt.next;
prev = prev.next;
}
k = k % total;
prev.next = head;
int count = 0;
while (count < total - k){
prev = prev.next;
count++;
}
dummy.next = prev.next;
prev.next = null;
return dummy.next;
}
}
