61. 旋转链表

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61. 旋转链表

官方链接

https://leetcode-cn.com/problems/rotate-list/

给定一个链表，旋转链表，将链表每个节点向右移动 k 个位置，其中 k 是非负数。

示例 1:

输入: 1->2->3->4->5->NULL, k = 2
输出: 4->5->1->2->3->NULL
解释:
向右旋转 1 步: 5->1->2->3->4->NULL
向右旋转 2 步: 4->5->1->2->3->NULL

示例 2:

输入: 0->1->2->NULL, k = 4
输出: 2->0->1->NULL
解释:
向右旋转 1 步: 2->0->1->NULL
向右旋转 2 步: 1->2->0->NULL
向右旋转 3 步: 0->1->2->NULL
向右旋转 4 步: 2->0->1->NULL

解法一

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode rotateRight(ListNode head, int k) {

        if (head == null || k == 0) return head;

        ListNode dummy = new ListNode(-1);
        dummy.next = head;

        ListNode pre = dummy;
        pre = pre.next;

        int len = 0;
        while (pre != null) {
            pre =  pre.next;
            len++;
        }

        k %= len; 
        if (k == 0) return head;

        ListNode slow = dummy, fast = dummy;

        while (k-- > 0) fast = fast.next;

        while (fast.next != null) {
            slow = slow.next;
            fast = fast.next;
        }

        ListNode next = dummy.next;
        dummy.next = slow.next;
        slow.next = null;
        fast.next = next;

        return dummy.next;

    }
}