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 ->…

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 ->…

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 -&…

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. You may assume the two num…

我现在在做一个叫<leetbook>的免费开源书项目,力求提供最易懂的中文思路,目前把解题思路都同步更新到gitbook上了,需要的同学可以去看看 书的地址:https://hk029.gitbooks.io/leetbook/ ` 2.Add Two Numbers [M] Add Two Numbers M 题目 思路 代码 更精巧的代码 题目 You are given two linked lists representing two non-negative numbers. The…

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 numbers and return it as a linked list. You may assum…

题目: 两个数字求和,数字用链表表示,每一个结点代表一位.链表顺序与数字顺序相反,即表头存放数字的最低位. 解法: 分别遍历两个链表的每个结点,对两个结点求和即可.要维护一个变量保存每次相加之后的进位. 更常见的,链表顺序与数字顺序相同,那么做一次链表逆序,求和之后再逆序回来即可. 代码: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int…

[题目描述] You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in forward order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the…

1.问题描述 2.问题分析 循环拆分数字,然求和判断. 3.代码 int addDigits(int num) { ) return num; int result = num; do{ vector<int> r = splitnum( result ); result = ; for(auto & n : r ){ result += n; } } ); return result ; } vector<int> splitnum( int num ){ vector&…

1.题目描述 2.问题分析 直接按照加法运算规则运算即可,注意进位问题. 3.代码 string addStrings(string num1, string num2) { if( num1.empty() ) return num2; if( num2.empty() ) return num1; string::reverse_iterator it1 = num1.rbegin() ; string::reverse_iterator it2 = num2.rbegin() ; stri…