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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { int carry = 0; ListNode tail = new ListNode(0); ListNode ptr = tail; while(l1!=null || l2!=null) { int tmp1 = 0;//必须要有，避免长度不一样 if(l1!=null) { tmp1 = l1.val; l1 = l1.next; } int tmp2 = 0;//同上 if(l2!=null) { tmp2 = l2.val; l2 = l2.next; } int tmp = tmp1 + tmp2 + carry; carry = tmp/10; ptr.next = new ListNode(tmp%10); ptr = ptr.next; } //如果两个链表都为空了，但最后一位进了1的情况 if(carry==1) { ptr.next = new ListNode(1); } return tail.next; } }
原文：http://www.cnblogs.com/zhangmingzhao/p/7286507.html