题解 | #反转链表#

两个链表解决：

public class Solution { public ListNode ReverseList(ListNode head) { if(head == null) return null; //定义新链表 ListNode reverseList = null;

    while(head != null){
        //保存下一节点和后面的所有节点
        ListNode node = head.next;
        //最关键的，从reverseList的头部插入
        head.next = reverseList;
        //reverseList更新
        reverseList = head;
        //赋值，将改变后的head只想燕来的下一个节点
        head = node;
    }

    return reverseList;
    
}

}