被ListNode的结构玩儿死了,还是不够熟悉,怪自己
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 Explanation: 342 + 465 = 807.
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode list = new ListNode(0); ListNode q = list; int sum; boolean plus=false; while(l1!=null||l2!=null){ sum=0; if(plus==true){ sum+=1; plus=false; } if(l1!=null){ sum+=l1.val; l1=l1.next; } if(l2!=null){ sum+=l2.val; l2=l2.next; } if(sum>9){ plus=true; q.next=new ListNode(sum%10); }else{ plus=false; q.next=new ListNode(sum%10); } q=q.next; } if(plus==true){ q.next=new ListNode(1); q=q.next; } return list.next; } }
在里面其实有挺多坑的
1.在调用.next的时候,应该先确认这个节点是否有下一个节点
2.之前一直尝试着:ListNode result=new ListNode(0);(全局)
ListNode q=new ListNode(...)(局部),result=q;
现在已经搞不明白当时思路了:
而正确思路应该是:首先建立一个返回的节点,再建立一个临时节点变量,使得这个变量指向返回节点,然后对这个变量赋值下一个节点,等等
3.题目中说明了不用返回以0值开头的节点,一开始在想这怎么实现,其实很简单:先创建一个0值的头结点,在最后返回的时候,返回头结点的next下一个节点