题目来源：LeetCode 67. 二进制求和

//Java：二进制加法
public class JFETK5 {
    
    public static void main(String[] args) {
        Solution solution = new JFETK5().new Solution();
        // TO TEST
        System.out.println(solution.addBinary("101111","10"));
    }
    
    //leetcode submit region begin(Prohibit modification and deletion)
    class Solution {
        public String addBinary(String a, String b) {
            // 1.逆序字符串
            StringBuilder sa = new StringBuilder();
            StringBuilder sb = new StringBuilder();
            sa.append(a);
            sb.append(b);
            sa.reverse();
            sb.reverse();
            // 2.长度短的那个字符串补零
            int n = 0;
            if(sa.length() > sb.length()) {
                n = sa.length() - sb.length();
                for(int i = 0 ; i < n ; i++) {
                    sb.append('0');
                }
            } else if (sa.length() < sb.length()){
                n = sb.length() - sa.length();
                for(int i = 0 ; i < n ; i++) {
                    sa.append('0');
                }
            }
            // 3.逐位进行加法运算
            StringBuilder stringBuilder = new StringBuilder();
            int flag = 0;   // 判断是否有进位，有进位则为1，否则为0，不用boolean是为方便计算
            for(int i = 0 ; i < sa.length() && i < sb.length() ; i++) {
                if(flag == 0) { // 没有进位
                    if(sa.charAt(i) == sb.charAt(i) && sa.charAt(i) == '0') {
                        flag = 0;
                        stringBuilder.append('0');
                    } else if(sa.charAt(i) == sb.charAt(i) && sa.charAt(i) == '1') {
                        flag = 1;
                        stringBuilder.append('0');
                    } else {
                        flag = 0;
                        stringBuilder.append('1');
                    }
                } else if(flag == 1){   // 有进位
                    if(sa.charAt(i) == sb.charAt(i) && sa.charAt(i) == '0') {
                        flag = 0;
                        stringBuilder.append('1');
                    } else if(sa.charAt(i) == sb.charAt(i) && sa.charAt(i) == '1') {
                        flag = 1;
                        stringBuilder.append('1');
                    } else {
                        flag = 1;
                        stringBuilder.append('0');
                    }
                }
            }
            if(flag == 1) {
                stringBuilder.append('1');
            }
    
            // 逆序stringBuilder
            return stringBuilder.reverse().toString();
    
        }
    }
    //leetcode submit region end(Prohibit modification and deletion)
}