LeetCode 2. Add Two Numbers

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

给定两个链表，按顺序相加，并向后进位，合成并返回一个链表

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode l3 = null;        ListNode node = null;        for(int sum=0;l1!=null||l2!=null||sum!=0;sum/=10){            if(l1!=null){                sum += l1.val;                l1 = l1.next;            }            if(l2!=null){                sum += l2.val;                l2 = l2.next;            }            if(l3==null){                l3 = new ListNode(0);                l3.next = new ListNode(sum%10);                l3 = l3.next;                node = l3;            }            else{                node.next = new ListNode(sum%10);                node = node.next;            }        }        return l3;    }}

思路是用一个变量sum去累加l1,l2的值，并将对10取余得到的值传给新的链表，每次循环结束将sum除以10，若为1则是进位。