题解 | #反转链表#

public class Solution {
    public ListNode ReverseList(ListNode head) {
        ListNode cur = head;
        ListNode pre = null;
        while(cur!=null){
            ListNode curNext = cur.next; //下一个节点
            cur.next = pre;//当前节点的指针指向pre
            pre = cur;//pre更新为当前节点
            cur = curNext;//当前节点指向下一个节点,即前面存储的curNext
           
        }
        return pre;
    }
}