题解 | #判断链表中是否有环#

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    public ListNode oddEvenList (ListNode head) {
        // write code here
        if(head == null ) return head ;
        if(head.next == null) return head ;
        
        ArrayList<Integer> left = new ArrayList<>();
        ArrayList<Integer> right = new ArrayList<>();
        int cnt = 1;
        while (head !=null ){
            if (cnt % 2 == 1){
                left.add(head.val);
            }else{
                right.add(head.val);
            }
            head = head.next ;
            cnt++ ;
        }
        
        ListNode dummarynode = new ListNode(0);
        ListNode cur = dummarynode ;
        
        for(int i = 0 ;i < left.size();i++){
            cur.next = new ListNode(left.get(i));
            cur = cur.next ;
        }
        for(int i = 0 ;i < right.size();i++){
            cur.next = new ListNode(right.get(i));
            cur = cur.next ;
        }        
        return dummarynode.next ;
    }
}

定义左右两个数组用来存放链表中的数据，再将数据存放到链表当中。 时间复杂度O(n),空间复杂度也为O(n).