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# 655.Big Integer Addition

## 1.Description(Easy)

Given two non-negative integers`num1`and`num2`represented as string, return the sum of`num1`and`num2`.

### Notice

* The length of both num1 and num2 is < 5100.
* Both num1 and num2 contains only digits 0-9.
* Both num1 and num2 does not contain any leading zero.
* You must not use any built-in BigInteger library or convert the inputs to integer directly.

**Example**

Given num1 =`"123"`, num2 =`"45"`\
return`"168"`

[**Tags**](https://www.lintcode.com/en/problem/big-integer-addition/#tags)

[Mathematics](https://www.lintcode.com/tag/mathematics/) [Airbnb](https://www.lintcode.com/tag/airbnb/) [Google](https://www.lintcode.com/tag/google/)

## 2.Code

乘法的答案在这里：<http://www.jiuzhang.com/solutions/big-integer-multiplication/>

跟167 Add two number基本一样，需要一个carry控制进位。

**需要注意的是把char转换成int的时候要用num.charAt(i)-'0',而不能直接强制转换(int).**

```
public String addStrings(String num1, String num2) {
        if((num1==null ||num1.length()==0)&& (num2==null || num2.length()==0)){
            return "";
        }
        if(num1==null ||num1.length()==0){
            return num2;
        }
        if(num2==null || num2.length()==0){
            return num1;
        }

        int carry=0;
        StringBuilder result=new StringBuilder();
        int i=num1.length()-1;
        int j=num2.length()-1;
        while(i>=0 && j>=0){
            int sum=num1.charAt(i)-'0'+num2.charAt(j)-'0'+carry;
            carry=sum/10;
            int current=sum%10;
            result.append(current);
            i--;
            j--;
        }
        while(i>=0){
            int sum=num1.charAt(i)-'0'+carry;
            carry=sum/10;
            int current=sum%10;
            result.append(current);
            i--;
        }
        while(j>=0){
            int sum=num2.charAt(j)-'0'+carry;
            carry=sum/10;
            int current=sum%10;
            result.append(current);
            j--;
        }
        if(carry!=0){
            result.append(carry);
        }

        String res=result.reverse().toString();
        return res;

    }
```