LeetCode_ 2 (add two numbers)

Title Description ：
Here are two for you Non empty The linked list of , Represents two nonnegative integers . Each of them is based on The reverse Stored in , And each node can only store a Numbers .

Please add up the two numbers , And returns a linked list representing sum in the same form .

You can assume that in addition to the numbers 0 outside , Neither of these numbers 0 start .

Example 1：

Input ：l1 = [2,4,3], l2 = [5,6,4]
Output ：[7,0,8]
explain ：342 + 465 = 807.

Example 2：
Input ：l1 = [0], l2 = [0]
Output ：[0]

Example 3：
Input ：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output ：[8,9,9,9,0,0,0,1]

Tips ：
The number of nodes in each list is in the range [1, 100] Inside
0 <= Node.val <= 9
The title data guarantees that the number indicated in the list does not contain leading zeros

/**
 * Definition for singly-linked list.
 * public class ListNode {
    
 *     int val;
 *     ListNode next;
 *     ListNode() {
    }
 *     ListNode(int val) {
     this.val = val; }
 *     ListNode(int val, ListNode next) {
     this.val = val; this.next = next; }
 * }
 */
class Solution {
    
   public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    
        ListNode head=null,l=null;      // Create head nodes and pointer nodes 
        int flag=0;                     // Carry mark 
        while(l1!=null||l2!=null){
          // Traverse two single linked lists 
            int n1=l1!=null?l1.val:0;   // Get the value of the current node of the first linked list , If the current node is empty , Then fill in zero 
            int n2=l2!=null?l2.val:0;   // ditto 
            int cur=n1+n2+flag;         // The value that should be filled in the linked list of calculation results 
            if(head==null){
                 // If it is the first value , Then create a node and connect the head pointer 
                l=new ListNode(cur%10);
                head=l;
            }else{
                          // Create a new node, fill in the result and move the pointer back 
                l.next=new ListNode(cur%10);
                l=l.next;
            }
            flag=cur/10;                // Judge whether carry is needed 
            if(l1!=null)l1=l1.next;     // Judge before moving the pointer back , To avoid pointer exceptions 
            if(l2!=null)l2=l2.next;
        }
        if(flag>0)l.next=new ListNode(flag);
        return head;
    }
}