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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

注意：add函数的第一个参数tmp是指针的指针。刚开始用的是指针，结果出错，原因如下。ret的值为0x005af618，在main作用域tmp也等于0x005af618。到了add()函数里，tmp的值虽然被赋值为next=0x005af650，但是当返回main函数的时候，仍变成0x005af618。指针就是地址，也是值传参的，符合作用域。这个是我之前的一个误会，以为只要指针肯定会被改变值并带着返回。

另外，最开始报错地址不对齐，是因为没有给next指针初始化为NULL。

#include <stdlib.h>
/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     struct ListNode *next;
* };
*/
inline int add(struct ListNode** tmp,int ll1,int ll2,int carry){
struct ListNode* next;
next = (struct ListNode*) malloc(sizeof(struct ListNode));
next->val = carry ? (ll1 + ll2 + carry) % 10 : (ll1 + ll2)%10;
next->next = NULL;
carry = carry ? (ll1 + ll2 + carry) / 10 : (ll1 + ll2)/10;
(*tmp)->next = next;
*tmp = next;
return carry;
}
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
struct ListNode* ll1 = l1;
struct ListNode* ll2 = l2;
struct ListNode* ret;
ret = (struct ListNode*) malloc(sizeof(struct ListNode));
int carry = 0;
ret->val = (ll1->val + ll2->val) % 10;
ret->next = NULL;
carry = (ll1->val + ll2->val) / 10;
ll1 = ll1->next;
ll2 = ll2->next;
struct ListNode* tmp = ret;
while (ll1 && ll2) {
carry = add(&tmp,ll1->val,ll2->val,carry);
ll1 = ll1->next;
ll2 = ll2->next;
}
while (ll1) {
carry = add(&tmp,ll1->val,0,carry);
ll1 = ll1->next;
}
while (ll2) {
carry = add(&tmp,ll2->val,0,carry);
ll2 = ll2->next;
}
while(carry!=0){
carry = add(&tmp,0,0,carry);
}
return ret;
}