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 numbers do not contain any leading zero, except the number 0 itself.
Example:
Input : ( 2 -> 4 -> 3 ) + ( 5 -> 6 -> 4 )
Output：7-> 0 -> 8
Explanation：342 + 465 = 807
思路：
example 1:
2->4->3
5->6->4
(2+5+0)->(4+6+0)->(3+4+1) 第三个数代表进位
7->0->8
example 2:
1
9->9
(1+9+0)->(0+9+1)->(0+0+1)
0->0->1
代码：
/** * 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, ListNode* l2) { ListNode *res = l1; // the res is result ListNode *cur = l1; // record the last ListNode of the first singly-linked list int carry = 0; while( l1 != nullptr && l2 != nullptr ){ // if l1 and l2 are not nullptr,then add them up int bitSum = l1->val + l2->val + carry; carry = bitSum / 10; bitSum = bitSum % 10; l1->val = bitSum; cur = l1; l1 = l1->next; l2 = l2->next; } if( l2 != nullptr ){ // if l2 is not empty , link l1 and l2 cur->next = l2; l1 = cur->next; } while( l1 != nullptr ){ int bitSum = l1->val + carry; carry = bitSum / 10; bitSum = bitSum % 10; l1->val = bitSum; cur = l1; l1 = l1->next; } if( carry != 0 ){ ListNode *t = new ListNode( carry ); cur->next = t; } return res; } };
转载请标明出处:LeetCode 2 : Add Two Numbers