链表表示的两数相加

给定两个非空链表来表示两个非负整数。位数按照逆序方式存储，它们的每个节点只存储单个数字。将两数相加返回一个新的链表。你可以假设除了数字 0 之外，这两个数字都不会以零开头。

示例：

输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)

输出：7 -> 0 -> 8

原因：342 + 465 = 807

代码一：

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode result = new ListNode(0);
        ListNode first = l1, second = l2, tmpNode = result;
        int tmp = 0;
        
        while(first != null || second != null){
            int firstValue ,secondValue;
            if(first == null){firstValue = 0;}
            else{firstValue = first.val;}
            
            if(second == null){secondValue = 0;}
            else{secondValue = second.val;}
            
            tmp = tmp + firstValue + secondValue;
            tmpNode.next = new ListNode(tmp % 10);
            tmp = tmp / 10;
            
            tmpNode = tmpNode.next;
            if(first != null){first = first.next;}
            if(second != null){second = second.next;}
        }
        if(tmp != 0){
            tmpNode.next = new ListNode(tmp % 10);
        }
        return result.next;
    }

代码二：

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummyHead = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummyHead;
    int carry = 0;
    while (p != null || q != null) {
        int x = (p != null) ? p.val : 0;
        int y = (q != null) ? q.val : 0;
        int sum = carry + x + y;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummyHead.next;
}