Rotate List —— LeetCode

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;
    }
}