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
思路：乍一看以为是大整数，实则相当简单，比Add Binary还要简单
代码：
1 ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { 2 return addTwoDigit(l1,l2,0); 3 } 4 ListNode *addTwoDigit(ListNode *d1, ListNode *d2, int carry){ 5 int sum = carry; 6 ListNode *n1 = d1, *n2 = d2;//不要轻易修改输入参数, 尤其是链表问题中 7 if(d1 != NULL){ 8 sum += d1->val; 9 n1 = d1->next; 10 } 11 if(d2 != NULL){ 12 sum += d2->val; 13 n2 = d2->next; 14 } 15 ListNode *newNode = new ListNode(sum%10); 16 17 if(d1 == NULL && d2 == NULL){ 18 if(sum == 0) 19 return NULL;//Avoid test case:{0}, {0} => {0,0} 20 return newNode; 21 } 22 23 ListNode *nextNode = addTwoDigit(n1, n2, sum/10); 24 newNode->next = nextNode; 25 return newNode; 26 }