题解 | #反转链表#

比较简单的一道题，要义就是拿住当前、上一个节点、下一个节点，保持不断链就可以了。

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode ReverseList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode cur = head;
        ListNode last = null;
        ListNode next = cur.next;
        while (cur != null) {
            next = cur.next;
            cur.next = last;
            last = cur;
            cur = next;
        }
        return last;
    }
}