这个题目主要是需要处理两个数长度不一样的情况，有两个解决办法。我是采用递归，把长的那一段与carray再加。也可以把短的前面补0.
public class Solution { public String addBinary(String a, String b) { int p1 = a.length() - 1; int p2 = b.length() - 1; int carry = 0; String result = ""; while (p1 >= 0 && p2 >= 0) { int one = a.charAt(p1) - ‘0‘; int two = b.charAt(p2) - ‘0‘; result = String.valueOf(one ^ two ^ carry) + result; carry = carry + one + two > 1 ? 1 : 0; p1 --; p2 --; } if (p1 >= 0) result = addBinary(String.valueOf(carry), a.substring(0, p1 + 1)) + result; else if (p2 >= 0) result = addBinary(String.valueOf(carry), b.substring(0, p2 + 1)) + result; else result = carry == 1? String.valueOf(carry) + result : result; return result; } }
原文：http://www.cnblogs.com/vision-love-programming/p/5011837.html