python,List保存节点Val

Python版本

class Solution:
    def reverseBetween(self , head , m , n ):
        # write code here
        res=[]
        while head:
            res.append(head.val)//此处理论可以保存head，但实际会超时，此题目不太严谨，仅仅通过交换节点的val，就能通过
            head=head.next      //测试案例，而题目本意，是反转区间节点而不是反转val值

        res=res[0:m-1]+res[m-1:n][::-1]+res[n:]

        first=ListNode(0)
        head1=first
        for i in res:
            temp=ListNode(i)
            head1.next=temp
            head1=temp
        return first.next

java版本

import java.util.*;
import java.util.Iterator;
/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param head ListNode类 
     * @param m int整型 
     * @param n int整型 
     * @return ListNode类
     */
    public ListNode reverseBetween (ListNode head, int m, int n) {
        // write code here
        List<Integer> list=new ArrayList<>();
        ListNode head1=head;
        while(head1!=null){
            list.add(head1.val);
            head1=head1.next;
        }
       List<Integer> tt=new ArrayList<>();
        tt.addAll(list.subList(0,m-1));
        tt.addAll(rev(list.subList(m-1,n)));
        tt.addAll(list.subList(n,list.size()));
        Iterator iter=list.iterator();
        ListNode first=new ListNode(0);
        ListNode head2=first;
        while(iter.hasNext()){
            ListNode temp=new ListNode((int)iter.next());
            head2.next=temp;
            head2=temp;
        }
        return first.next;
    }
    public static List rev(List list){
        Collections.reverse(list);
        return list;
    }
}