[JAVA]用数学解释数组移步问题(新增对链表移步的解释）

数组移步转载：https://blog.csdn.net/shijing_0214/article/details/53056943

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题目来源：leetcode 2019-03-17

题目描述：Given a linked list, rotate the list to the right by k places, where k is non-negative.

链表移步：观察链表移步的规律可以发现，其实就是在一个环里走，然后选一个结点当头就行

Input: 0->1->2->NULL, k = 4
Output: 2->0->1->NULL
Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right: 0->1->2->NULL
rotate 4 steps to the right: 2->0->1->NULL

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代码如下：


/**
 * 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(k==0||head==null) return head;
        ListNode cur=head;
        int l=1;
        while(cur.next!=null)//链表长度
        {
            ++l;
            cur=cur.next;
        }
        if(k>l) k=k%l;
        if(l==k) return head;//步数如果是整数倍直接返回
        cur.next=head;//成环
        cur=head;
        k=l-k;//找头结点
        while(--k!=0)
        {
            cur=cur.next;
        }
        head=cur.next;
        cur.next=null;
        return head;
    }
    
}