LeetCode No.2 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

初看这题，就是C1课上学的超长整数加减法。于是思路就很明确了：

c_n = (a_n + b_n + carry) % 10

carry = (a_n + b_n + carry) / 10

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        int carry = 0, cn, tmp;
        ListNode *c = NULL, *p, *q, *r, *t;

        p = l1; q = l2; r = c;

        while (p && q) {
            tmp = p->val + q->val + carry;
            cn = tmp % 10;
            carry = tmp / 10;　　　　　　　　　　　　　　  // Add t to end of result list.
            t = new ListNode(cn);
            if (c == NULL)
                c = t;
            else
                r->next = t;
            r = t;

            p = p->next; q = q->next;
        }

        t = p ? p : q;
        　　     // When p or q is not entirely processed, keep calculating.
        while (t) {
            tmp = t->val + carry;
            cn = tmp % 10;
            carry = tmp / 10;
            r->next = new ListNode(cn);
            r = r->next;
            t = t->next;
        }
　　　　　　　　　　 // Finally, handle case like 99999 + 1
        if (carry) {
            r->next = new ListNode(carry);
        }

        return c;
    }

运行时间比较理想，163ms