leetcode——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
要注意边做进位加法边创建链表，一开始一直用node=node->next;这样链表不存在，不能得出结果，并考虑到最后一位还有进位的情况
（ps：第一次做leetcode，如果有时间还要回来看看，刷了10多天终于做到Medium了，要加快进度）

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode *head = (ListNode *)malloc(sizeof(ListNode));
ListNode *pre = head;
ListNode *node = NULL;
//进位
int c = 0,sum;
//加法
while(l1 != NULL && l2 != NULL){
sum = l1->val + l2->val + c;
c = sum / 10;
node = (ListNode *)malloc(sizeof(ListNode));
node->val = sum % 10;
node->next = NULL;
//尾插法
pre->next = node;
pre = node;
l1 = l1->next;
l2 = l2->next;
}
//例如：2->4->3->1   5->6->4
while(l1 != NULL){
sum = l1->val + c;
c = sum / 10;
node = (ListNode *)malloc(sizeof(ListNode));
node->val = sum % 10;
node->next = NULL;
//尾插法
pre->next = node;
pre = node;

l1 = l1->next;
}
//例如：2->4->3   5->6->4->1
while(l2 != NULL){
sum = l2->val + c;
c = sum / 10;
node = (ListNode *)malloc(sizeof(ListNode));
node->val = sum % 10;
node->next = NULL;
//尾插法
pre->next = node;
pre = node;
l2 = l2->next;
}
//最后一位还有进位
if(c > 0){
node = (ListNode *)malloc(sizeof(ListNode));
node->val = c;
node->next = NULL;
//尾插法
pre->next = node;
pre = node;
}
return head->next;

}
};