两数相加

题目来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/add-two-numbers

给出两个 非空 的链表用来表示两个非负的整数。其中，它们各自的位数是按照 逆序 的方式存储的，并且它们的每个节点只能存储 一位 数字。

如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。

您可以假设除了数字 0 之外，这两个数都不会以 0 开头。

示例：

输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
原因：342 + 465 = 807

Java代码：

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

Python代码：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        head = ListNode(0)
        curr = ListNode(0)
        head.next = curr
        while l1 is not None or l2 is not None:
            if l1 is None:
                curr.next = l2
                break
            if l2 is None:
                curr.next = l1
                break

            sum = l1.val + l2.val

            l1 = l1.next
            l2 = l2.next

            if sum < 10:
                curr.next = ListNode(sum)
            else:
                curr.next = ListNode(sum%10)
                if l1 is None and l2 is None:
                    l1 = ListNode(1)
                elif l1 is None:
                    l1 = ListNode(1)
                elif l2 is None:
                    l2 = ListNode(1)
                else:
                    l1.val = l1.val + 1

            curr = curr.next
        return head.next.next