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;

    }

}