文章作者：Tyan
博客：noahsnail.com  |  CSDN  |  简书

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;
    }
}