C# 寫 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.


代碼：

static void Main(string[] args)
        {
            ListNode l1 = new ListNode(3);
            l1.next = new ListNode(6);
            l1.next.next = new ListNode(9);

            ListNode l2 = new ListNode(7);
            l2.next = new ListNode(0);
            l2.next.next = new ListNode(3);

            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(0);
            ListNode head = l3;
            int sum = 0;
            while (l1 != null || l2 != null)
            {
                sum = sum > 9 ? 1 : 0;
                if (l1 != null)
                {
                    sum += l1.val;
                    l1 = l1.next;
                }
                if (l2 != null)
                {
                    sum += l2.val;
                    l2 = l2.next;
                }
                //存儲在l3中
                l3.next = new ListNode(sum % 10);
                l3 = l3.next;
            }
            //判斷最後一項是否和大於9，大於則須要再添加一個1.
            if (sum > 9)
            {
                l3.next = new ListNode(1);
            }
            return head.next;
        }

 

解析：

輸入：ListNode類型的兩個參數

輸出：第一個節點。

思想：

　　循環鏈表中的每一位，sum存儲兩個鏈表對應位上的和。經過觀察不難發現規律，若是上一位和大於9，則下一位初始sum爲1，將結果存儲在新的鏈表中。

　　最後一位上和大於9時，再多加一位，值爲1。

時間複雜度：O(n)