LeetCode OJ：Add Binary（二进制相加）

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

For example,
a = "11"
b = "1"
Return "100".

简单的二进制相加而已，只不过传入的参数是字符串而已。为了方便，先将string  reverse了一下，代码如下：

 class Solution {
 public:
     string addBinary(string a, string b) {
         reverse(a.begin(), a.end());
         reverse(b.begin(), b.end());
         int len1 = a.length();
         int len2 = b.length();
         string result;
         int flag ,val;
         flag = val = ;
         int i;
         for(i = ; i < len1 && i < len2; ++i){
             val = (a[i] - '') + (b[i] - '') + flag;
             result.append(, val %  + '');
             flag = val/;
         }
         while (i < len1) {
             val = a[i] - ''+ flag;
             result.append(, val %  + '');
             flag = val/;
             ++i;
         }
         while (i < len2 ){
             val = b[i] - ''+ flag;
             result.append(, val %  + '');
             flag = val/;
             ++i;
         }
         cout << "flag " << flag << endl;
         if(flag != )
             result.append(, '');
         reverse(result.begin(), result.end());
         return result;
     }
 };

java：以前的循环用的好蠢啊，居然用了三次循环，下面的java稍微有些改进：

 public class Solution {
     public String addBinary(String a, String b) {
         int len1 = a.length();
         int len2 = b.length();
         String ret = new String();
         int i = 0;
         int carry = 0;
         a = ReverseStr(a);
         b = ReverseStr(b);
         while(i < len1 || i < len2 || carry != 0){
             int val = ((i<len1)?(a.charAt(i)-'0'):0) + ((i<len2)?(b.charAt(i)-'0'):0) + carry;
             carry = 0;
             if(val > 1){
                 carry = val/2;
                 ret += (char)((val%2) + '0');
             }else{
                 ret += (char)(val + '0');
             }
             i++;
         }
         return ReverseStr(ret);
     }

        public String ReverseStr(String str){
         return new StringBuffer(str).reverse().toString();
     }
 }