Q2：Add Two Numbers

2. Add Two Numbers

官方的链接：2. Add Two Numbers

Description ：

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 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.

---

问题描述

给定2个非空的代表非负整数的链表，数字是倒过来存储的，用链表表示求和。可以假设没有前导数字0。

思路

使用迭代解法，注意最后的进位。

[github-here]

 public class Q2_AddTwoNumbers {

     /**
      * 迭代解法
      *
      * @param ListNode
      *            l1
      * @param ListNode
      *            l2
      * @return
      */
     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
         if (null == l1 && null == l2) {
             return new ListNode(0);
         }
         // 保存链表头，便于创建结点和最后返回结果
         ListNode headNode = new ListNode(0);
         ListNode sumNode = headNode;
         // 和以及进位
         int sum = 0;
         int carry = 0;
         while (null != l1 && null != l2) {
             sum = l1.val + l2.val + carry;
             sumNode.next = new ListNode(sum % 10);
             carry = sum / 10;
             sumNode = sumNode.next;
             l1 = l1.next;
             l2 = l2.next;
         }
         while (null != l1) {
             sum = l1.val + carry;
             sumNode.next = new ListNode(sum % 10);
             carry = sum / 10;
             sumNode = sumNode.next;
             l1 = l1.next;
         }
         while (null != l2) {
             sum = l2.val + carry;
             sumNode.next = new ListNode(sum % 10);
             carry = sum / 10;
             sumNode = sumNode.next;
             l2 = l2.next;
         }
         if (carry > 0) {
             sumNode.next = new ListNode(carry);
             sumNode = sumNode.next;
         }
         return headNode.next;
     }
 }