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); l3 = l3.next; node = l3; } else{ node.next = new ListNode(sum); 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。