题解 | #反转链表#

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

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode ReverseList(ListNode head) {
        //特殊值（空指针）处理
        if(head == null){
            return null;
        }
        
        ListNode prev = null;//当前指针的前向节点
        ListNode next = head;//当前指针的后向节点
        while(head != null){
            next = head.next;//保存当前节点的后向节点的引用
            head.next = prev;//更新当前指针的后向节点
            prev = head;//更新当前指针的前向节点
            head = next;//更新当前节点
        }
        
        return prev;//返回反转链表的头指针
    }
}