# 67. Add Binary / 415. Add Strings

[ **67. Add Binary**](https://leetcode.com/problems/add-binary/submissions/1)

Given two binary strings, return their sum (also a binary string).

The input strings are both **non-empty** and contains only characters `1` or `0`.

**Example 1:**

```
Input: a = "11", b = "1"
Output: "100"
```

**Example 2:**

```
Input: a = "1010", b = "1011"
Output: "10101"
```

**My Solutions:**

新建StringBuilder, 从a和b的末尾开始加，如果结果是1记成carry。最后加上carry，反转sb。

Time: O(n)

```
public class Solution {
    public String addBinary(String a, String b) {
        StringBuilder sb = new StringBuilder();
        int i = a.length() - 1, j = b.length() - 1, carry = 0;
        while (i >= 0 || j >= 0) {
            int sum = carry;
            if (i >= 0) sum += a.charAt(i--) - '0';
            if (j >= 0) sum += b.charAt(j--) - '0';
            sb.append(sum % 2);
            carry = sum / 2;
        }
        if (carry != 0) sb.append(carry);
        return sb.reverse().toString();
    }
}
```

[ **415. Add Strings**](https://leetcode.com/problems/add-strings/description/)

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

**Note:**

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

**My Solutions:**

和add binary类似，binary 用%2， 此题用%10&#x20;

```
public class Solution {
    public String addStrings(String num1, String num2) {
        
        StringBuilder sb = new StringBuilder();
        int carry = 0;
        for(int i = num1.length() - 1, j = num2.length() - 1; i >= 0 || j >= 0 || carry == 1; i--, j--){
            int x = i < 0 ? 0 : num1.charAt(i) - '0';
            int y = j < 0 ? 0 : num2.charAt(j) - '0';
            sb.append((x + y + carry) % 10);
            carry = (x + y + carry) / 10;
        }
        return sb.reverse().toString();
        
    }
}
```