2. Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.


首先想到能不能根据小学数学中两整数相加的过程，实现下。思考了下程序该怎么做。发现可以。

考虑几种情况，第一个链表的长度大于，小于，等于第二个链表，在最后末尾有无溢出。开始写下程序。

这里存在一个决定，是该选择第一个链表，还是选择第二个链表存储最终的结果。最好的选择是存在长的链表里，这样malloc申请的内存会少一些，减少空间占用。可是想要得到链表长度，需要n个步骤。
所以我就随机选了第一个链表，链表1大于链表2的概率是50%，我至少有一半的概率选到长的链表，而不需要计算两者的长度。我现在还不能下判断，到底哪种更好。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    int overflow = 0, tmp;
    struct ListNode *p1,*p2;
    p1 = l1;
    p2 = l2;
    int i=1;
    
    //process the one before the last one of common length
    while(p1->next && p2->next) {
            tmp = p1->val + p2->val + overflow;
            overflow = tmp / 10;
            p1->val = tmp % 10;
            
            p1 = p1->next;
            p2 = p2->next;
    }
    
    
    if (!p1->next) { //when length of l1 <= l2
        tmp = p1->val + p2->val + overflow;
        overflow = tmp / 10;
        p1->val = tmp % 10;
        
        //when l1 < l2 and no overflow.
        if(!overflow && p2->next) {
            p1->next = p2->next;
        }
        
        while (overflow) {
            if (!p2->next){
                p2->next = malloc(sizeof(struct ListNode));
                memset(p2->next, 0, sizeof(struct ListNode));
            }
            if(i==1) p1->next = p2->next;
            i++;
            p2 = p2->next;
            tmp = p2->val + overflow;
            overflow = tmp / 10;
            p2->val = tmp % 10;
        }
    } else {//when legngth of l1 > l2
        tmp = p1->val + p2->val + overflow;
        overflow = tmp / 10;
        p1->val = tmp % 10;
        
        while (overflow) {
            if (!p1->next) {
                p1->next = malloc(sizeof(struct ListNode));
                memset(p1->next,0,sizeof(struct ListNode));
            }
            p1 = p1->next;
            tmp = p1->val + overflow;
            overflow = tmp / 10;
            p1->val = tmp % 10;
        }
    }
    
    return l1;
    
}