You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example 1:
Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 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]
思路：
这个问题的解决方案实际上就是模拟了我们手动相加两个数的过程。首先，我们创建一个空链表来存储每一位的相加结果。在每一步中，我们将对应的节点值相加，并加上可能存在的进位（carry）。然后，我们将相加的结果对10取余数，这个余数就是新的节点的值。最后，我们通过将相加的结果除以10，得到的商就是下一步的进位。如果最后还有进位，我们会在链表的末尾添加一个新的节点
答案：
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
head := &ListNode{}
current , n1 ,n2 , carray := head, 0,0,0
for l1 != nil || l2 != nil || carray != nil{
if l1 == nil{
n1 = 0
}else{
n1 = l1.Val
l1 = l1.Next
}
if l2 == nil{
n2 = 0
}else{
n2 = l2.Val
l2 = l2.Next
}
current.Next = &ListNode{Val:(n1 + n2 +carray)%10}
carray := (n1 + n2 +carray) /10
}
return current.Next
}