LeetCode--Add Two Numbers

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#Add Two Numbers
##题目
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.

##分析
在本题中，基本的思路很简单，就是以最长的那条链为加法链，那么较短的链的高位如果不存在的话那么我们就当其值为0，还有就是要考虑到进位，最高位要进位的时候要申请空间。我遇到的最大的一个坑就是

ListNode *result = new ListNode(0);

本来我在这里只是申请了一个变量，即

ListNode *result;

这时返回的链就会是空的。
猜测是没有申请空间的话地址是不确定的，所以会有这样的情况出现。

##源码

/**
 * 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 *result = new ListNode(0);
        ListNode *re = result;
        int sum = 0, carry = 0;
        while(l1 != NULL||l2 != NULL) {
        	int num1,num2;
        	if(l1 == NULL) {
        		num1 = 0;
        	} else {
        		num1 = l1->val;
        	}
        	if(l2 == NULL) {
        		num2 = 0;
        	} else {
        		num2 = l2->val;
        	}
        	sum = num1 + num2 + carry;
        	result->next = new ListNode(sum%10);
        	result = result->next;
        	carry = sum/10;
        	if(l1 != NULL) {
        		l1 = l1->next;
        	}
        	if(l2 != NULL) {
        		l2 = l2->next;
        	}
        }
        if(carry != 0){
        	result->next = new ListNode(carry);
 		}
 		return re->next;    
	}
};