Add Two Numbers 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/problems/add-two-numbers/description/ Description 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 c…

LeetCode(2) || Add Two Numbers && Longest Substring Without Repeating Characters 题记 刷LeetCode第二天,今天做了两道题,速度比我想的要慢,看样子两个月的目标有点难,尽力吧.今天主要做了Add Two Numbers和Longest Substring Without Repeating Characters 两道题,下面就来看下这两道题吧. 题一:Add Two Numbers 题目内容 You ar…

题目: LeetCode:1. Add Two Numbers 描述: Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice.…

You are given two linked lists representing two non-negative numbers. The most significant digit comes first 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 con…

1 题目 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 digit. Add the two numbers and return it as a linked list. Input: (2 -> 4 -> 3) + (5 -> 6 -…

索引 思路1:基本加法规则 思路2:移花接木法... 问题描述:https://leetcode.com/problems/add-two-numbers/ 思路1:基本加法规则 根据小学学的基本加法规则.....我们需要将两个数以最低位为基准对齐,然后逐个加,需要进位的给进位就行了....恰好这个链表是逆序的!!!已经为我们对齐了.用两个指针分别指向两个链表头,开始同步往后移动,边移动边计算结果,直到两个指针都到了尽头/没有进位.时间复杂度是O(max(m, n)),空间复杂度和时间复杂度一样…

[LeetCode] Add Two Numbers 两个数字相加   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…

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…

Add Two Numbers 方法一: 考虑到有进位的问题,首先想到的思路是: 先分位求总和得到 totalsum,然后再将totalsum按位拆分转成链表: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { ; ; while(l1 != NULL && l2 != NULL) { sum += i*(l1->val + l2->val); i *= ; l1 = l1->next; l2 = l2->n…

add two numbers 看题一脸懵逼,看中文都很懵逼,链表怎么实现的,点了debug才看到一些代码 改一下,使本地可以跑起来 # Definition for singly-linked list. class ListNode: def __init__(self, x): self.val = x self.next = None class Solution: def addTwoNumbers(self, l1, l2): """ :type l1: List…