[转载请注明]http://www.cnblogs.com/igoslly/p/8672467.html 来看一下题目: 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 a…

1.题目描述 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…

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 -&…

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…

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…

反向存储,从左往右加 ［抄题］: 你有两个用链表代表的整数,其中每个节点包含一个数字.数字存储按照在原来整数中相反的顺序,使得第一个数字位于链表的开头.写出一个函数将两个整数相加,用链表形式返回和.给出两个链表 3->1->5->null 和 5->9->2->null,返回 8->0->8->null [暴力解法]: 时间分析: 空间分析: ［思维问题］: 直接从左往右加就行,没看出来. 链表结构改变时要添加dummy,返回的是dummy.next…

题意:两个非空链表求和,这两个链表所表示的数字没有前导零,要求不能修改原链表,如反转链表. 分析:用stack分别存两个链表的数字,然后从低位开始边求和边重新构造链表. Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 8 -> 0 -> 7 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListN…