算法学习（Java实现之字符串篇）····二进制求和

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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
//若字符串a的长度小于字符串b的长度，两个字符串交换
if(a.length() < b.length()){
String temp = a;
a =b;
b = temp;
}
int la = a.length()-1;
int lb = b.length()-1;
int carries = 0;//进位
String res = "";//结果值
while(lb >= 0){//先依据短字符串的长度依次计算
int sum = (int)(a.charAt(la)-'0')+(int)(b.charAt(lb)-'0')+carries;
res = String.valueOf(sum%2)+res;
carries =sum/2;
la--;
lb--;
}
while(la>=0){//再依据长字符串的长度依次计算
int sum = (int)(a.charAt(la)-'0')+carries;
res = String.valueOf(sum%2)+res;
carries =sum/2;
la--;
}

if (carries==1){
res ="1"+res;//计算最后的进位
}
return res;
}
}