# Leetcode-067 - Binary Sum

#### Binary summation

Description: Give you two binary strings and return their sum (in binary).

The input is a non-empty string and contains only the digits 1 and 0.

See the LeetCode website for an example.

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###### Solution 1: group processing

ParseInt (String, 2) is used to convert the binary number of a String into a decimal int. This method is used to convert the binary number of a String into a decimal int. This method is used to convert the binary number of a String into a decimal int. The integer.tobinaryString (int) method is then used to convert the added value back to a binary string, and the pieced strings are added up to produce the final result.

There are several points to consider in the process of doing this:

- First, when adding each segment, consider whether there is a carry. If there is a carry, use addOne to represent the carry number with a value of 1, and add addOne to the next segment.
- Second, when the length of each segment is less than 30 digits, it is necessary to add 0 in front of it to 30 digits.

`public class LeetCode_067 { public static String addBinary(String a, String b) { int addOne = 0; String result = ""; while (a.length() 30 || b.length() 30) { String aStr; String bStr; if (a.length() 30) { aStr = a.substring(a.length() - 30, a.length()); a = a.substring(0, a.length() - 30); } else { aStr = a; a = "0"; } if (b.length() 30) { bStr = b.substring(b.length() - 30, b.length()); b = b.substring(0, b.length() - 30); } else { bStr = b; b = "0"; } String temp = Integer.toBinaryString(Integer.parseInt(aStr, 2) + Integer.parseInt(bStr, 2) + addOne); if (temp.length() 30) { addOne = 1; temp = temp.substring(1, temp.length()); } else { if(temp.length() 30) { int zeroCount = 30 - temp.length(); for(int i = 0; i zeroCount; i++) { temp = "0" + temp; } } addOne = 0; } result = temp + result; } return Integer.toBinaryString(Integer.parseInt(a, 2) + Integer.parseInt(b, 2) + addOne) + result; } public static void main(String[] args) { String a = "1001101011011010000010111010100111001000100001111110011111001010100101111"; String b = "111000011000010000001100001001010011000101000000001111101101000100000000100100001100010000111001000"; System.out.println(addBinary(a, b)); }}`

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