AI代码解释 publicstaticListNodeaddTwoNumbers2(ListNode l1,ListNode l2){// 边界条件判断if(l1==null){returnl2;}elseif(l2==null){returnl1;}ListNode list=null;ListNode next=null;// 记录和值int sum=0;// 记录是否有进位int b=0;while(l1!=null||l2!=null){if(l1!=null){sum=l1.val;l1=l1...
以前我认为归并需要分成3部分,while()直到node1和node2其中一个为空,然后再while()非空的那个node,但是不同场景应用方式也不同,如果遇到归并相加问题,就不用那么麻烦,代码对比一下就知道了。 publicclassSolution {publicstaticvoidmain(String[] args) {int[] data1 =newint[] {1};int[] data2 =newint[]...
LeetCode---Add Two Numbers 解法 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 ...
val=0, next=None): # self.val = val # self.next = next class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: link1 = l1 link2 = l2 while(link1!=None and link2 !=None): a = link1.val link2.val += a link1.val = link2.val link1 = link1....
首发于LeetCode 每日一题 切换模式写文章 登录/注册LeetCode 2 - Add Two Numbers | 两数相加 (Python3|Go) 满赋诸机 前小镇做题家,现大厂打工人。 题意 给定两个非空链表表示的不含前导零的非负整数(逆序存储),求这两个整数的和,并以相同形式的链表返回。 数据限制 两个链表中的结点数在 [1, 100] ...
leetcode算法—两数相加 Add Two Numbers 关注微信公众号:CodingTechWork,一起学习进步。 题目 Add Two Numbers: 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 ...
public class Solution { /* * 方法1 */ public static ListNode addTwoNumbers(ListNode l1,ListNode l2) { //如果都为空 直接返回不为空的一个参数 如果都未空 则返回空 if(l1 == null || l2 == null){ return l1 == null ?(l2 == null ?null:l2):l1; ...
3 输入与输出：/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode(int x) : val(x), next(NULL) {}* };*/class Solution {public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { }};4 解决思路：从表头开始相加，记录每次相加...
方法一:转成数字后相加运算求和,将和转为字符串后再转为List # Definition for singly-linked list.# class ListNode(object):# def __init__(self, x):# self.val = x# self.next = NoneclassSolution(object):defaddTwoNumbers(self,l1,l2):""" :type l1: ListNode :type l2: ListNode :rtype:...
(node4);ListNode*result2=Solution().addTwoNumbers(node3,node4);printNode(result2,true);printf("---\n");free(node1);free(node2);free(node3);free(node4);free(result1);free(result2);}voidtestSpec(){ListNode*node1=ListNode::nodeWithNumber(0);ListNode*node2=ListNode::nodeWithNumber(...