其實仔細想想是挺簡單的，我們要做的只是記得進位。

LinkedListNode addLists(LinkedListNode l1, LinkedListNode l2, int carry)
//LinkedListNode addLists(LinkedListNode l1, LinkedListNode l2, int carry = 0)
{
    if (l1 == null && l2 == null $$ carry == 0)
    {
        return null;
    }
    
    LinkedListNode result = new
 
 LinkedListNode();
    int value = carry;
    if (l1 != null)
    {
        value += l1.data;
    }
    if (l2 != null)
    {
        value += l2.data;
    }
    
    result.data = value%10;
    
    LinkedListNode more = addLists(l1 == null? null : l1.next, l2 == null? null : l2.next, value >= 10? 1 :0);
    result.setNext(more);
}
 

進階：假設是正向存放的。

坑：1，註意雙方長度，雙方是末尾對齊的，不足的地方需要用0補足。

//2.5b進階：假設這些數位是正向存放的
public class PartialSum
{
    public LinkedListNode sum = null;
    public int carry = 0;
}

LinkedListNode addLists(LinkedListNode l1, LinkedListNode l2)
{
    int len1 = length(l1);
    int len2 = length(l2);
    
    //用零填充較短的鏈表
    if
 
 (len1 < len2)
    {
        l1 = padList(l1, len2 - len1);
    }
    else
    {
        l2 = padList(l2, len1 - len2);
    }
    
    //對兩個鏈表求和
    PartialSum sum = addListsHelper(l1, l2);
    
    //如有進位，則插入鏈表首部，否則，直接返回整個鏈表
    if (sum.carry == 0)
    {
        return sum.sum;
    }
    else
    {
        LinkedListNode result = insertBefore(sum.sum, sum.carry);
        return result;
    }
}

PartialSum addListsHelper(LinkedListNode l1, LinkedListNode l2)
{
    if (l1 == null && l2 == null)
    {
        PartialSum sum = new PartialSum();
        return sum;
    }
    
    //先遞歸為較小數字求和
    PartialSum sum = addListsHelper(l1.next, l2.next);
    
    //將進位和當前數據相加
    int val = sum.carry + l1.data + l2.data;
    
    //插入當前數字的求和結果
    LinkedListNode full_result = insertBefore(sum.sum, val % 10);
    
    //返回求和結果與進位值
    sum.sum = full_result;
    sum.carry = val/10;
    return sum;
}

//用零填充鏈表
LinkedListNode padList(LinkedListNode l,int padding)
{
    LinkedListNode head = l;
    for (int i = 0; i < padding; i++)
    {
        LinkedListNode n = new LinkedListNode(0, null, null)；
        head.prev = n;
        n.next = head;
        head = n;
    }
    return head;
}

//輔助函數，將結點插入鏈表首部
LinkedListNode insertBefore(LinkedListNode list, int data)
{
    LinkedListNode node = new LinkedListNode(data, null, null);
    if (list != null)
    {
        list.prev = node;
        node.next = list;
    }
    return node;
}

2.5給定兩個用鏈表表示的整數，每個結點包含一個數位。這些數位是反向存放的，也就是個位排在鏈表首部。編寫函數對這兩個整數求和，並用鏈表形式返回結果。