LeetCode - 445 - Add Two Numbers II

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up:

What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

这个题，一直RE，RE到怀疑人生，RE到又做了几题回来重写了几次，还是RE   orzzzzzzzzzz

RE的问题是，我在while循环里写了如下的语句。

ListNode node(sum % 10);
ListNode* cur = &node;

new的内存区域在堆，但是node临时变量开在栈区，这么写导致出现了double free问题，所以不停地RE。double free是指程序分配一块内存之后，经过使用将这块内存释放，但并没有将指向这块内存的所有指针抹零或回收，并在其他部分再次将指向同一块内存单元的指针交给内存分配器去进行释放操作。

/**
* 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) {
if (!l1) return l2;
if (!l2) return l1;
stack<ListNode*> s1, s2;
while (l1) {
s1.push(l1);
l1 = l1->next;
}
while (l2) {
s2.push(l2);
l2 = l2->next;
}
int sum = 0;
ListNode* pre = NULL;
ListNode* cur = NULL;
while (!s1.empty() || !s2.empty() || sum) {
if (!s1.empty()) {
sum += s1.top()->val;
s1.pop();
}
if (!s2.empty()) {
sum += s2.top()->val;
s2.pop();
}
cur = new ListNode(sum % 10);
cur->next = pre;
pre = cur;
sum = sum / 10;
}
return pre;
}
};