LintCode-408.二进制求和

public class Solution {
    /**
     * @param a a number
     * @param b a number
     * @return the result
     */
    public String addBinary(String a, String b) {
        // Write your code here
        StringBuffer sb = new StringBuffer();
        char carry = '0';
        int count;
        char[] as = a.toCharArray();
        char[] bs = b.toCharArray();
        char[] longer = (as.length > bs.length) ? as : bs;
        int minLength = (as.length > bs.length) ? bs.length : as.length;
        for (int i = 0; i < minLength; i++){
            count = add(as[as.length - i - 1], bs[bs.length - i - 1], carry);
            if(count == 0) {carry = '0';sb.append('0');}
            else if(count == 1) {carry = '0';sb.append('1');}
            else if(count == 2) {carry = '1';sb.append('0');}
            else if(count == 3) {carry = '1';sb.append('1');}
            else return "0";
        }
        for(int j = minLength; j < longer.length; j++){
            count = add(longer[longer.length - j - 1], carry, '0');
            if(count == 0) {carry = '0';sb.append('0');}
            else if(count == 1) {carry = '0';sb.append('1');}
            else if(count == 2) {carry = '1';sb.append('0');}
            else return "0";
        }
        if(carry == '1') sb.append('1');
        return sb.reverse().toString();
    }
    public int add(char a, char b, char c){
        short count = 0;
        count += ((a == '0') ? 0 : 1);
        count += ((b == '0') ? 0 : 1);
        count += ((c == '0') ? 0 : 1);
        return count;
    }
}