/usr/bin/python32#-*- coding: utf-8 -*-3#@author: Albert4#@Time: 2018/8/13567classListNode(object):8def__init__(self, x):9self.val =x10self.next =None1112classSolution(object):13defaddTwoNumbers(self, l1, l2):#求解14"""15:type l1: ListNode16:type l2: ListNode17:rtype: Lis...
4.2个数相加,可能会产生最高位的进位,因此要注意在完成以上1-3的操作后,判断进位是否为0,不为0则需要增加结点存储最高位的进位。 给个链接[LeetCode]Add Two Numbers 原来是倒着相加,342+465=807,结果倒序,正好是708,题目应该是这样理解的吧,哈哈。 本题的思路很简单,按照小学数学中学习的加法原理从末尾到首位...
Leetcode 解题 Add Two Numbers Python 原题:You are given two linked lists representing two non-negative numbers.The digits are stored in reverse order and each of their nodes contain a single digitAdd the two numbers and return it as a linked listInput: (2->4->3) + (5->6->4)Output...
Add Two Numbers 二、解题 1)题意 给出两个链表,把对应位置上的值进行十进制相加(有进位),返回链表的根节点。 2)输入输出说明 输入:两个列表的根节点(并不是整个列表,即leetcode会把默认生成好的列表的根节点传入) 输出:累加之后的根节点 3)关键点 1)做十进制加法时,使用和%10来得出当前位的值,使用/10...
代码(Python3) # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: # 哨兵结点,方便后续处...
leetcode 传送门 例子: Input:(2->4->3)+(5->6->4)Output:7->0->8Explanation:342+465=807. 大体意思就是从前向后做加法,逢十进一。 在评论区找到的一个代码: classSolution(object):defaddTwoNumbers(self,l1,l2):""" :type l1: ListNode ...
【LeetCode】2. Add Two Numbers 两数相加 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 个人公众号:负雪明烛 本文关键词:两数相加,链表,求加法,题解,leetcode, 力扣,python, c++, java 目录 题目描述
Python 代码如下: # Definition for singly-linked list. # class ListNode(object): # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution(object): def addTwoNumbers(self, l1, l2): st1 = [] ...
【LeetCode题解---2】Add Two Numbers 题目 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....
*/publicclassSolution{/** * @param l1: the first list * @param l2: the second list * @return: the sum list of l1 and l2 */publicListNodeaddLists(ListNode l1,ListNode l2){// write your code here// write your code here// save the pre nodeListNode pre=newListNode(0);// save the...