[LeetCode] Add Two Numbers 链表】的更多相关文章

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 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…
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 ->…
Add Two Numbers这个问题的意思是,提供两条链表,每条链表表示一个十进制整数,其每一位对应链表的一个结点.比如345表示为链表5->4->3.而我们需要做的就是将两条链表代表的整数加起来,并创建一个新的链表并返回,新链表代表加总值. 这个问题与算法无关,其主要是操作链表这一数据结构,以及两个大数的加法.由于没有规定链表的长度,这意味着链表代表的整数可以任意大,这样就不能先利用链表计算出对应的整数值,之后利用加总值重新生成链表.我们必须手动处理两个大数的加法. 要谨慎两个个位数的加总…
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…
原题链接在这里:https://leetcode.com/problems/add-two-numbers-ii/ 题目: 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…
Add Two NumbersYou 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…
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 -&…
数字反过来这个没有什么麻烦,就是镜像的去算十进制加法就可以了,然后就是简单的链表. /** * 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, ListN…
注意进位的处理和节点为null的处理 public ListNode addTwoNumbers(ListNode l1, ListNode l2) { int flag = 0; ListNode res = null; ListNode temp = null; //注意如果进位不是0的话还要添加一个节点 while (l1!=null||l2!=null||flag!=0) { int cur = flag; flag = 0; if (l1!=null) { cur+=l1.val; l…