【题解】【链表】【Leetcode】Add 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 -> 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：

 ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {

     return addTwoDigit(l1,l2,);

 }

 ListNode *addTwoDigit(ListNode *d1, ListNode *d2, int carry){

     int sum = carry;

     ListNode *n1 = d1, *n2 = d2;//不要轻易修改输入参数, 尤其是链表问题中

     if(d1 != NULL){

         sum += d1->val;

         n1 = d1->next;

     }

     if(d2 != NULL){

         sum += d2->val;

         n2 = d2->next;

     }

     ListNode *newNode = new ListNode(sum%);

     if(d1 == NULL && d2 == NULL){

         if(sum == )

             return NULL;//Avoid test case:{0}, {0} => {0,0}

         return newNode;

     }

     ListNode *nextNode = addTwoDigit(n1, n2, sum/);

     newNode->next = nextNode;

     return newNode;

 }

巴特西

【题解】【链表】【Leetcode】Add Two Numbers

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