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 -> 4)
Output: 7 -> 0 -> 8
这道并不是什么难题，算法很简单，链表的数据类型也不难。就是建立一个新链表，然后把输入的两个链表从头往后撸，每两个相加，添加一个新节点到新链表后面，就是要处理下进位问题。还有就是最高位的进位问题要最后特殊处理一下。代码如下：
C++ 解法：
class Solution { public: ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { ListNode *res = new ListNode(-1); ListNode *cur = res; int carry = 0; while (l1 || l2) { int n1 = l1 ? l1->val : 0; int n2 = l2 ? l2->val : 0; int sum = n1 + n2 + carry; carry = sum / 10; cur->next = new ListNode(sum % 10); cur = cur->next; if (l1) l1 = l1->next; if (l2) l2 = l2->next; } if (carry) cur->next = new ListNode(1); return res->next; } };
Java 解法：
public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode dummy = new ListNode(-1); ListNode cur = dummy; int carry = 0; while (l1 != null || l2 != null) { int d1 = l1 == null ? 0 : l1.val; int d2 = l2 == null ? 0 : l2.val; int sum = d1 + d2 + carry; carry = sum >= 10 ? 1 : 0; cur.next = new ListNode(sum % 10); cur = cur.next; if (l1 != null) l1 = l1.next; if (l2 != null) l2 = l2.next; } if (carry == 1) cur.next = new ListNode(1); return dummy.next; } }
本文转自博客园Grandyang的博客，原文链接：两个数字相加[LeetCode] Add Two Numbers ，如需转载请自行联系原博主。
