Leetcode-Rotate List

最新推荐文章于 2024-05-15 22:45:44 发布

转载最新推荐文章于 2024-05-15 22:45:44 发布 · 50 阅读

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原文链接：http://www.cnblogs.com/lishiblog/p/4102796.html

本文介绍了一种链表右旋算法，即给定一个链表，将其向右旋转k个位置，其中k为非负数。例如，对于链表1->2->3->4->5->NULL和k=2，返回4->5->1->2->3->NULL。文章分析了当k大于链表长度时的情况，并提供了一个解决方案。

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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.

Have you met this question in a real interview?

Analysis:

Assume the length of the list is len, we need to consider the situation that k > len which means to rotate the right k%n steps. Since rotate operation is a kind of circling operation. We first count the length of the list, and connect end to head to make the list as a circle. Then, we only need to move forward by len-k%len steps to reach the new end of the list. Now the next node is the new head.

Solution:

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public ListNode rotateRight(ListNode head, int n) {
14         if (head==null || head.next==null) return head;
15     
16         ListNode end = null;
17         int len = 1;
18         end = head;
19         while (end.next!=null){
20             end = end.next;
21             len++;
22         }
23         end.next = head;
24         int step = len - n%len;
25         int index = 1;
26         end = head;
27         while (index<step){
28             end = end.next;
29             index++;
30         }
31         head = end.next;
32         end.next = null;
33         return head;              
34     }
35 }

转载于:https://www.cnblogs.com/lishiblog/p/4102796.html