leetcode2

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {
        var list = new List<ListNode>();
            var temp1 = l1;
            var temp2 = l2;
            var step = 0;
            while (l1 != null || l2 != null)
            {
                var v1 = 0;
                var v2 = 0;
                if (l1 == null)
                {
                    l1 = new ListNode(0);
                }
                v1 = l1.val;
                l1 = l1.next;

                if (l2 == null)
                {
                    l2 = new ListNode(0);
                }
                v2 = l2.val;
                l2 = l2.next;

                var cur = v1 + v2 + step;
                var result = 0;
                if (cur >= 10)
                {
                    step = 1;
                    result = cur % 10;
                }
                else
                {
                    step = 0;
                    result = cur;
                }
                list.Add(new ListNode(result));
            }
            if (step == 1)
            {
                list.Add(new ListNode(1));
            }

            for (int i = 0; i < list.Count - 1; i++)
            {
                list[i].next = list[i + 1];
            }
            var head = list[0];
            return head;
    }
}

https://leetcode.com/problems/add-two-numbers/#/description

补充python实现：

class Solution:
    def addTwoNumbers(self, l1: 'ListNode', l2: 'ListNode') -> 'ListNode':
        headnode = ListNode(0)
        temp = headnode
        carry = 0
        while  l1!=None or l2!=None:
            if l1==None:
                l1 = ListNode(0)
            if l2==None:
                l2 = ListNode(0)
            cur = l1.val + l2.val + carry
            if cur // 10 > 0:
                carry = 1
            else:
                carry = 0
            cur = cur % 10
            temp.next = ListNode(cur) 
            temp = temp.next
            l1 = l1.next
            l2 = l2.next
        if carry > 0:
            temp.next = ListNode(carry)
        return headnode.next