leetcode61旋转链表

Posted 2021-09-18 lisin-lee-cooper

一.问题描述

给你一个链表的头节点 head ，旋转链表，将链表每个节点向右移动 k 个位置。
输入：head = [1,2,3,4,5], k = 2
输出：[4,5,1,2,3]

二.示例代码

public class RotatingLinkedList61 {

    public static void main(String[] args) {

        ListNode listNode1 = new ListNode(1);
        ListNode listNode2 = new ListNode(2);
        ListNode listNode3 = new ListNode(3);
        ListNode listNode4 = new ListNode(4);
        ListNode listNode5 = new ListNode(5);

        listNode1.next = listNode2;
        listNode2.next = listNode3;
        listNode3.next = listNode4;
        listNode4.next = listNode5;

        int k = 2;

        ListNode result = rotatingLinkedList(listNode1, k);
        System.out.println(result);

    }

    private static ListNode rotatingLinkedList(ListNode root, int k) {
        int length = 1;
        ListNode tempNode = root;
        while (tempNode.next != null) {
            length++;
            tempNode = tempNode.next;
        }
        k %= length;
        int moveIndex = length - k;
        tempNode.next = root;
        while (moveIndex > 0) {
            tempNode = tempNode.next;
            moveIndex--;
        }
        ListNode result = tempNode.next;
        tempNode.next = null;
        return result;
    }
    
}