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文章作者：Tyan

1. 问题描述

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

2. 求解

方法一

先求解两个链表的和，直接一个链表结束或两个链表同时结束，然后再处理没结束链表的剩下部分。

public class Solution {
       public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        int sum = 0;        int a = 0;        int b = 0;        int quotient = 0;        ListNode head = new ListNode(0);        ListNode current = head;        while(l1 != null && l2 !=null) {            a = l1.val;            b = l2.val;            sum = a + b + quotient;            current.next = new ListNode(sum % 10);            quotient = sum / 10;            l1 = l1.next;            l2 = l2.next;            current = current.next;        }        ListNode temp = null;        if(l1 != null) {            temp = l1;        }else if(l2 != null) {            temp = l2;        }else {            if(quotient != 0) {                temp = new ListNode(0);            }        }        while(temp != null) {            sum = temp.val + quotient;            current.next = new ListNode(sum % 10);            quotient = sum / 10;            temp = temp.next;            current = current.next;        }        if(quotient != 0) {             current.next = new ListNode(quotient);        }        return head.next;    }}

方法二

方法一中的代码有较多的冗余，例如current.next = new ListNode(sum % 10);出现了两次，两次while循环的逻辑是非常类似的，经过代码的变换可以将两部分合成一部分，即同时处理两个链表直至两个链表都结束。

public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        int sum = 0;        int a = 0;        int b = 0;        int quotient = 0;        ListNode head = new ListNode(0);        ListNode current = head;        while(l1 != null || l2 !=null) {            if(l1 ==null) {                a = 0;            }else {                a = l1.val;                l1 = l1.next;            }            if(l2 == null) {                b = 0;            }else {                b = l2.val;                l2 = l2.next;            }            sum = a + b + quotient;            current.next = new ListNode(sum % 10);            quotient = sum / 10;            current = current.next;        }        if(quotient != 0) {             current.next = new ListNode(quotient);        }        return head.next;    }}

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