【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,l2,0);
}
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%10);
    
    if(d1 == NULL && d2 == NULL){
        if(sum == 0)
            return NULL;//Avoid test case:{0}, {0} => {0,0}
        return newNode;
    }
    
    ListNode *nextNode = addTwoDigit(n1, n2, sum/10);
    newNode->next = nextNode;
    return newNode;
}