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# Leetcode 2 Add Two Numbers 题解分析

发表于 2020-10-11 分类于 Java ， leetcode ， java ， linked list ， linked list 阅读次数： 阅读次数： Disqus：

又 roll 到了一个以前做过的题，不过现在用 Java 也来写一下，是 easy 级别的，所以就简单说下

## 简要介绍

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 contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.
就是给了两个链表，用来表示两个非负的整数，在链表中倒序放着，每个节点包含一位的数字，把他们加起来以后也按照原来的链表结构输出

## 样例

#### example 1

    Input: l1 = [2,4,3], l2 = [5,6,4]
    Output: [7,0,8]
    Explanation: 342 + 465 = 807.

#### example 2

    Input: l1 = [0], l2 = [0]
    Output: [0]

#### example 3

    Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
    Output: [8,9,9,9,0,0,0,1]

## 题解

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
            ListNode root = new ListNode();
            if (l1 == null && l2 == null) {
                return root;
            }
            ListNode tail = root;
            int entered = 0;
            // 这个条件加了 entered，就是还有进位的数
            while (l1 != null || l2 != null || entered != 0) {
                int temp = entered;
                if (l1 != null) {
                    temp += l1.val;
                    l1 = l1.next;
                }
                if (l2 != null) {
                    temp += l2.val;
                    l2 = l2.next;
                }
                entered = (temp - temp % 10) / 10;
                tail.val = temp % 10;
                // 循环内部的控制是为了排除最后的空节点
                if (l1 != null || l2 != null || entered != 0) {
                    tail.next = new ListNode();
                    tail = tail.next;
                }
            }
    //        tail = null;
            return root;
        }

这里唯二需要注意的就是两个点，一个是循环条件需要包含进位值还存在的情况，还有一个是最后一个节点，如果是空的了，就不要在 new 一个出来了，写的比较挫