标签：lin bsp int ext head NPU || div efi
Given a linked list, rotate the list to the right by k places, where k is non-negative.
Example 1:
Input: 1->2->3->4->5->NULL, k = 2 Output: 4->5->1->2->3->NULL Explanation: rotate 1 steps to the right: 5->1->2->3->4->NULL rotate 2 steps to the right: 4->5->1->2->3->NULL
Example 2:
Input: 0->1->2->NULL, k = 4 Output:2->0->1->NULLExplanation: 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->NULLrotate 4 steps to the right:2->0->1->NULL
/** * 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 (head == null || head.next == null) { return head; } ListNode cur = head; int len = 1; while (cur.next != null) { cur = cur.next; len += 1; } cur.next = head; // use len - k%len for (int i = 1; i < len - k % len; i++) { head = head.next; } ListNode res = head.next; head.next = null; return res; } }
标签：lin bsp int ext head NPU || div efi
原文地址：https://www.cnblogs.com/xuanlu/p/12128516.html
