第30题 Add Binary

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

For example,

a = "11"

b = "1"

Return "100".

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Math String

Solution in Java:

public class Solution {
    public String addBinary(String a, String b) {
        int lengthA = a.length();
        int lengthB = b.length();
        int minLength, maxLength;
        String minStr, maxStr;
        if(lengthA<=lengthB){
            minLength = lengthA;
            minStr = a;
            maxLength = lengthB;
            maxStr = b;
        }else{
            minLength  = lengthB;
            minStr = b;
            maxLength = lengthA;
            maxStr = a;
        }
        String sum="";
        boolean carry = false;
        for(int i=0; i<minLength; i++){
            int digi1 = minStr.charAt(minLength-i-1)-'0';
            int digi2 = maxStr.charAt(maxLength-i-1)-'0';
            int sumDigi;
            if(carry)   sumDigi = digi1+digi2+1;
            else    sumDigi = digi1+digi2;
            if(sumDigi/2>0) carry = true;
            else    carry = false;
            sumDigi = sumDigi%2;
            String curDigi = String.valueOf(sumDigi);
            sum = curDigi +sum;
        }
        for(int i=maxLength-minLength-1; i>=0; i--){
            int digi = maxStr.charAt(i)-'0';
            if(carry) digi = digi+1;
            if(digi==2){
                digi=0;
                carry = true;
            }
            else    carry = false;
            String curDigi = String.valueOf(digi);
            sum = curDigi+sum;
        }
        if(carry)   sum = "1"+sum;
        return  sum;
    }
}

Note:

本来想直接用java中二进制，十进制和String间的转换，函数是：

二进制转十进制：int i=Integer.parseInt(str, 2)
第二个参数为进制树

十进制数转二进制String：String binStr = Integer.toBinaryString(i)

但在此题中不行，因为两个输入参数可能超过int和long的表示范围，会发生NumberFormatException，只能直接在String上操作。

获取String某一位的字符为str.charAt(i)，得到char的int值，可以用这个char减去‘0’，可以得到该位的0或1的值。char可以直接做减法。

int转为String用函数String str= String.valueOf(i)

注意这个加法操作要从String的末尾到开头进行