二进制求和

11

1

100

class Solution {
public:
/**
* @param a a number
* @param b a number
* @return the result
*/
string addBinary(string& a, string& b) {
// Write your code here
string result;
int temp = 0;
auto length = a.size() > b.size() ? a.size() : b.size();
reverse(a.begin(),a.end());
reverse(b.begin(),b.end());
for (auto i = 0; i < length; i++)
{
int aa = i < a.size() ? a[i] - '0' : 0;
int bb = i < b.size() ? b[i] - '0' : 0;
int val = (aa + bb + temp) % 2;
temp = (aa + bb + temp) / 2;
result.insert(result.begin(), val + '0');
}
if (temp == 1)
{
result.insert(result.begin(), 1 + '0');
}
return result;
}
};