Add Two Numbers

题目描述：

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

　　起初，这道题目被我想简单了。我用两个int值分别保存两个链表所表示的值，然后计算它们的和，最后把结果转换为链表。然而链表所能表示的值可以无限大，超出了int的存储范围。

后来自己思索了下，写出了以下代码：

solution1：

struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
    int first = 0;
    int second = 0;
    int carry = 0;
    int sum = 0;
    ListNode *list = new ListNode(-1);
    ListNode *p = list;
    while (l1 && l2)
    {
        first = l1->val;
        second = l2->val;
        sum = (first + second + carry) % 10;
        if(first + second + carry >= 10)
            carry = 1;
        else
            carry = 0;
        p->next = new ListNode(sum);
        p = p->next;
        l1 = l1->next;
        l2 = l2->next;
    }
    while (l1)
    {
        first = l1->val;
        if (carry)
        {
            sum = (first + carry) % 10;
            if(first + carry >= 10)
                carry = 1;
            else
                carry = 0;
        }
        else
        {
            sum = first;
        }
        p->next = new ListNode(sum);
        p = p->next;
        l1 = l1->next;
    }
    while (l2)
    {
        second = l2->val;
        if (carry)
        {
            sum = (second + carry) % 10;
            if(second + carry >= 10)
                carry = 1;
            else
                carry = 0;
        }
        else
        {
            sum = second;
        }
        p->next = new ListNode(sum);
        p = p->next;
        l2 = l2->next;
    }
    if (carry)
    {
        p->next = new ListNode(1);
    }
    return list->next;
}

代码比较冗余，且可读性不高。后来在论坛上找到了别人的优化算法，真的很赞。
solution2：

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
    ListNode *head = new ListNode(0);
    ListNode *p = head;
    int sum = 0;
    while (l1 != NULL || l2 != NULL)
    {
        if (l1 != NULL)
        {
            sum += l1->val;
            l1 = l1->next;
        }
        if (l2 != NULL)
        {
            sum += l2->val;
            l2 = l2->next;
        }
        p->next = new ListNode(sum % 10);
        p = p->next;
        sum /= 10;
    }
    if(sum)
        p->next = new ListNode(1);
    return head->next;
}

 

原文链接：https://oj.leetcode.com/discuss/2308/is-this-algorithm-optimal-or-what
ps:

　　代码AC只是第一步，要精益求精！