C# 写 LeetCode Medium #2 Add Two Numbers

2. Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse orderand each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

代码：

static void Main(string[] args)
        {
            ListNode l1 = new ListNode();
            l1.next = new ListNode();
            l1.next.next = new ListNode();

            ListNode l2 = new ListNode();
            l2.next = new ListNode();
            l2.next.next = new ListNode();

            var res= addTwoNumbers(l1, l2);
            while (res != null)
            {
                Console.Write(res.val);
                res = res.next;
            }
            Console.ReadKey();
        }

        public class ListNode
        {
            public int val;
            public ListNode next;
            public ListNode(int x) { val = x; }
        }

        public static ListNode addTwoNumbers(ListNode l1, ListNode l2)
        {
            ListNode l3 = new ListNode();
            ListNode head = l3;
            int sum = ;
            while (l1 != null || l2 != null)
            {
                sum = sum >  ?  : ;
                if (l1 != null)
                {
                    sum += l1.val;
                    l1 = l1.next;
                }
                if (l2 != null)
                {
                    sum += l2.val;
                    l2 = l2.next;
                }
                //存储在l3中
                l3.next = new ListNode(sum % );
                l3 = l3.next;
            }
            //判断最后一项是否和大于9，大于则需要再添加一个1.
            if (sum > )
            {
                l3.next = new ListNode();
            }
            return head.next;
        }

解析：

输入：ListNode类型的两个参数

输出：第一个节点。

思想：

　　循环链表中的每一位，sum存储两个链表对应位上的和。通过观察不难发现规律，如果上一位和大于9，则下一位初始sum为1，将结果存储在新的链表中。

　　最后一位上和大于9时，再多加一位，值为1。

时间复杂度：O(n)