67#二进制求和（6ms）（简单）

题目：

代码：

public String addBinary(String a, String b) {
                StringBuilder sb = new StringBuilder();
        String str_max = (a.length() >= b.length()) ? a:b;
        String str_min = (a.length() < b.length()) ? a:b;
        int count = 0;
        boolean flag = false;//判断是否进了一位
        while (count < str_min.length()){
            char num1 = str_min.charAt(str_min.length()-1-count);
            char num2 = str_max.charAt(str_max.length()-1-count);
            if (num1=='0' && num2=='0' && !flag){
                sb.append(0);
                flag = false;
            }else if (num1=='0' && num2=='0' && flag){
                flag = false;
            }else if (num1=='1' && num2=='1' && flag){
                sb.append(1);
            }else if(num1=='1' && num2=='1' && !flag){
                sb.append(0);
                sb.append(1);
                flag = true;
            } else{
                if (sb.length()!=0&&sb.charAt(sb.length()-1)=='1'&&flag){
                    sb.deleteCharAt(sb.length()-1);
                    sb.append(0);
                    sb.append(1);
                }else {
                    sb.append(1);
                    flag = false;
                }
            }
            count++;
        }
        if (a.length()!=b.length()) {
            while (count < str_max.length()){
                if (sb.charAt(sb.length()-1)=='1'&&str_max.charAt(str_max.length()-1-count)=='1'&&flag){
                    sb.deleteCharAt(sb.length()-1);
                    sb.append(0);
                    sb.append(1);
                }else if (sb.charAt(sb.length()-1)=='1'&&str_max.charAt(str_max.length()-1-count)=='0'&&flag){
                    flag = false;
                }else {
                    sb.append(str_max.charAt(str_max.length()-1-count));
                    flag = false;
                }
                count++;
            }
        }
        return sb.reverse().toString();
    }