[LeetCode]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

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
	public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
		if(l1 == null&&l2 == null){
			return null;
		}
		ListNode n1=l1;
		ListNode n2 =l2;
		ListNode node = new ListNode(0);
		ListNode head = node;
		int sum = 0;
		while(n1 !=null||n2 != null){
			if(n1!=null){
				sum+=n1.val;
				n1=n1.next;
			}
			
			if(n2!=null){
				sum+=n2.val;
				n2=n2.next;
			}
			
			node.next = new ListNode(sum%10);
			node=node.next;
			sum=sum/10;
		}
		
		if(sum==1){
			node.next = new ListNode(1);;
		}
		return head.next;
	}
}

class Solution {
public:
	ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
		ListNode head(-1);
		ListNode *cursor = &head;
		int tp =0;
		while(l1!=nullptr||l2!=nullptr){
			int x = l1!=nullptr?l1->val:0;
			int y = l2!=nullptr?l2->val:0;
			tp += x+y;
			cursor->next = new ListNode(tp%10);
			//不能返回局部对象的饮用或指针，函数完成后占用的存储空间也随之释放掉。
			//ListNode ln(tp%10);
			//cursor->next = &ln;
			cursor= cursor->next;
			if(l1!=nullptr) l1 = l1->next;
			if(l2!=nullptr) l2 = l2->next;
			tp /= 10;
		}
		if(tp!=0){
			cursor->next = new ListNode(tp);
		}
		return head.next;
	}
};