/**

* 思路：先直接相加，保留0，1，2等原值。再对所得的字符串按是否大于等于2来向前进位。

* AC：Runtime: 0 ms, faster than 100.00% of C++ online submissions for Add Binary.

* Memory Usage: 8.9 MB, less than 56.36% of C++ online submissions for Add Binary.

*

*/

class Solution {

public:

string addBinary(string a, string b) {

vector<char> avec, bvec;

for (int i = 0; i < a.length(); i++)

avec.push_back(a[i]);

for (int i = 0; i < b.length(); i++)

bvec.push_back(b[i]);

int range = avec.size() < bvec.size() ? avec.size() : bvec.size();

if (avec.size() > bvec.size())

swap(avec, bvec);

for (int i = range - 1; i >= 0; i--) {

bvec[bvec.size() - (range - i)] = char(int(bvec[bvec.size() - (range - i)] - int('0')) + int(avec[i]));

}

for (int i = bvec.size() - 1; i >= 0; i--) {

if (int(bvec[i]) >= int('2')) { // '2', '3' 转换成 '0', '1'

bvec[i] = char(int(bvec[i]) - 2);

if (i != 0) {

bvec[i - 1] = char(int(bvec[i - 1]) + 1);

}

else {

bvec.insert(bvec.begin(), '1');

}

}

}

string ans = "";

for (int i = 0; i < bvec.size(); i++)

ans += bvec[i];

return ans;

}

};