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# 167.Add Two Numbers

## 1.Description(Easy)

You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in`reverse`order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

**Example**

Given`7->1->6 + 5->9->2`. That is,`617 + 295`.

Return`2->1->9`. That is`912`.

Given`3->1->5`and`5->9->2`, return`8->0->8`.

[**Tags**](https://www.lintcode.com/en/problem/add-two-numbers/#tags)

[Cracking The Coding Interview](https://www.lintcode.com/tag/cracking-the-coding-interview/) [Linked List](https://www.lintcode.com/tag/linked-list/) [High Precision](https://www.lintcode.com/tag/high-precision/)

## 2.Code

从l1, l2头节点开始，对应位置相加并建立新节点。用一个变量carry记录进位。注意几种特殊情况：

1.一个链表为空

l1: NULL

l2: 1->2

sum: 1->2

2.l1, l2长度不同，且结果有可能长度超过l1, l2中的最大长度

l1: 2->2

l2: 9->9->9

sum: 1->2->0->1

```
public ListNode addLists(ListNode l1, ListNode l2) {

        if(l1==null && l2==null){
            return null;
        }

        ListNode dummy=new ListNode(0);
        ListNode current=dummy;
        int carry=0; //控制进位
        while(l1!=null && l2!=null){
            int sum=l1.val+l2.val+carry;//注意每次都要加上carry
            carry=sum/10;
            current.next=new ListNode(sum%10);

            l1=l1.next;
            l2=l2.next;
            current=current.next;
        }

        while(l1!=null){
            int sum=l1.val+carry;
            carry=sum/10;
            current.next=new ListNode(sum%10);

            l1=l1.next;
            current=current.next;
        }

        while(l2!=null){
            int sum=l2.val+carry;
            carry=carry/10;
            current.next=new ListNode(sum%10);

            l2=l2.next;
            current=current.next;
        }

        //最后判断下是否还有进位
        if(carry!=0){
            current.next=new ListNode(carry);
        }
        return dummy.next;
    }
```


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