本来是打算直接用 bitset 解决问题的，转成二进制再转成十进制，相加后再通过逆运算求得最终的字符串。
然而却存在一个问题，就是溢出，当给出的二进制过于大的时候，相加就解决不了了，因此就要按照字符串来处理。
以下是我的解决方案，虽说很丑，然而速度还行吧：
string addBinary(string a, string b) {
string result;
char carry = ‘0‘;
string* shortStrp = &a;
string* longStrp = &b;
if (a.length() > b.length()){
shortStrp = &b;
longStrp = &a;
}
auto pushFront = [&](const char* s)
{
result.insert(0, s);
};
auto longStrIt = longStrp->crbegin();
auto shortStrIt = shortStrp->crbegin();
while (longStrIt != longStrp->crend()){
if (shortStrIt == shortStrp->crend()){
auto const sum = (*longStrIt + carry);
if (sum == ‘0‘ + ‘0‘){
pushFront("0");
carry = ‘0‘;
}
else if (sum == ‘0‘ + ‘1‘){
pushFront("1");
carry = ‘0‘;
}
else if (sum == ‘1‘ + ‘1‘){
pushFront("0");
carry = ‘1‘;
}
++longStrIt;
}
else{
auto const sum = (*longStrIt + *shortStrIt + carry);
if (sum == ‘0‘ + ‘0‘ + ‘0‘){
pushFront("0");
carry = ‘0‘;
}
else if (sum == ‘0‘ +‘0‘ + ‘1‘){
pushFront("1");
carry = ‘0‘;
}
else if (sum == ‘1‘ + ‘1‘ + ‘0‘){
pushFront("0");
carry = ‘1‘;
}
else if (sum == ‘1‘ + ‘1‘ + ‘1‘){
pushFront("1");
carry = ‘1‘;
}
++longStrIt;
++shortStrIt;
}
}
if (carry == ‘1‘){
pushFront("1");
}
return result;
}
原文：http://www.cnblogs.com/wuOverflow/p/4783321.html