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。
1 public class Q2_AddTwoNumbers { 2 3 /** 4 * 迭代解法 5 * 6 * @param ListNode 7 * l1 8 * @param ListNode 9 * l2 10 * @return 11 */ 12 public ListNode addTwoNumbers(ListNode l1, ListNode l2) { 13 if (null == l1 && null == l2) { 14 return new ListNode(0); 15 } 16 // 保存链表头，便于创建结点和最后返回结果 17 ListNode headNode = new ListNode(0); 18 ListNode sumNode = headNode; 19 // 和以及进位 20 int sum = 0; 21 int carry = 0; 22 while (null != l1 && null != l2) { 23 sum = l1.val + l2.val + carry; 24 sumNode.next = new ListNode(sum % 10); 25 carry = sum / 10; 26 sumNode = sumNode.next; 27 l1 = l1.next; 28 l2 = l2.next; 29 } 30 while (null != l1) { 31 sum = l1.val + carry; 32 sumNode.next = new ListNode(sum % 10); 33 carry = sum / 10; 34 sumNode = sumNode.next; 35 l1 = l1.next; 36 } 37 while (null != l2) { 38 sum = l2.val + carry; 39 sumNode.next = new ListNode(sum % 10); 40 carry = sum / 10; 41 sumNode = sumNode.next; 42 l2 = l2.next; 43 } 44 if (carry > 0) { 45 sumNode.next = new ListNode(carry); 46 sumNode = sumNode.next; 47 } 48 return headNode.next; 49 } 50 }