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.

题目大意:给一个链表和一个非负整数k,把链表向右循环移位k位。

解题思路:踩了个坑,k有可能比链表的长度还长,比如长度为3的链表,k=2000000000,所以开始需要遍历一下链表算出长度,k%=len,然后就是两个pointer一个先走k步,另一个再走,最后把末尾的next节点设为head,新head就是慢指针指向的节点。

/**
* 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||k == 0){
return head;
}
int fast = k;
ListNode fastNode = head;
ListNode slowNode = head;
ListNode pre = slowNode;
int len = 0;
while(fastNode!=null){
len++;
fastNode=fastNode.next;
}
fast %= len;
fastNode = head;
if(fast == 0){
return head;
}
while(fastNode!=null&&fast-->0){
fastNode = fastNode.next;
}
while(fastNode!=null){
fastNode = fastNode.next;
pre = slowNode;
slowNode = slowNode.next;
}
ListNode newHead = new ListNode(0);
newHead.next = slowNode;
pre.next = null;
while(slowNode.next!=null){
slowNode = slowNode.next;
}
slowNode.next = head;
return newHead.next;
}
}