One brush 145 force deduction hot question-2 sum of two numbers (m)

 subject ：
 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 .
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 Example ：

 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 
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 reflection ：
 For the list problem , When the return result is the header node , Usually you need to initialize a pre pointer first  pre,
 The next node of the pointer points to the real header node head. The purpose of using the pre pointer is that there is no node value available when the list is initialized ,
 And the construction process of linked list needs pointer movement , This will result in the loss of the head pointer , Unable to return a result .

 label ： Linked list 
 Think of two linked lists as traversal of the same length , If a link list is short, fill it in the front  00,
 such as  987 + 23 = 987 + 023 = 1010
 Each bit needs to consider the carry problem of the previous bit at the same time , The carry value needs to be updated after the current bit calculation 
 If the two lists are all traversed , Carry value is  11, Then add the node at the front of the new list  11

/** 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) {
    
    	//res Store results ,cur by res The tail pointer of 
        ListNode res = new ListNode();
        ListNode cur = res;
        // It means carry 
        int carry = 0;
        while (l1 != null || l2 != null){
    
        	// If one of them reaches the end , Then the number of this digit in the linked list is 0.
            int a = l1 == null ? 0 : l1.val;
            int b = l2 == null ? 0 : l2.val;
            // The two digits of the two linked lists are added 
            int sum = a + b + carry;
            // Greater than 10 carry 
            carry = sum / 10;
            // The remainder after rounding 
            sum %= 10;
            // Create an access node res Back 
            cur.next = new ListNode(sum);
            cur = cur.next;
        	// Move back... Respectively 
            if (l1 != null) l1 = l1.next;
            if (l2 != null) l2 = l2.next;
        }
        // If there is a carry at the end , Add a node 
        if (carry == 1) cur.next = new ListNode(1);
        return res.next;
    }
}

LC