61. Rotate List

最新推荐文章于 2025-05-23 23:53:08 发布

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最新推荐文章于 2025-05-23 23:53:08 发布

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本文链接：https://blog.youkuaiyun.com/qq_39638957/article/details/94618528

本文深入探讨了链表右旋算法的实现细节，通过两个示例详细解释了如何将链表向右旋转k个位置，同时提供了完整的代码示例。文章首先给出了链表的基本定义，随后介绍了算法的具体步骤，包括确定链表长度、连接尾节点到头节点形成环、找到新的尾节点并断开链接，最后返回新的链表头节点。

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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->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

/**
 * 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) return head;
        
        int n = 1;
        ListNode tail = head;
        while(tail.next!=null){
            tail = tail.next;
            n++;
        }
        tail.next = head;
        
        ListNode newtail = head;
        for(int i = 0;i<n-k%n-1;i++){
            newtail = newtail.next;
        }
        ListNode newhead = newtail.next;
        newtail.next = null;
        
        return newhead;

    }
}