<p>Given two binary strings <code>a</code> and <code>b</code>, return <em>their sum as a binary string</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> a = "11", b = "1"
<strong>Output:</strong> "100"
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> a = "1010", b = "1011"
<strong>Output:</strong> "10101"
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= a.length, b.length &lt;= 10<sup>4</sup></code></li>
	<li><code>a</code> and <code>b</code> consist&nbsp;only of <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code> characters.</li>
	<li>Each string does not contain leading zeros except for the zero itself.</li>
</ul>

class Solution {
public:
    string addBinary(string a, string b) {
        string lstr = a;
        string sstr = b;
        if (a.size() > b.size()) {
            lstr = a;
            sstr = b;
        } else {
            lstr = b;
            sstr = a;
        }
        int adder = 0;
        int cond = 0;
        vector<int> ans;
        for (int i=0; i<lstr.size(); ++i) {
            if (i<sstr.size()) {
                cond = lstr[lstr.size()-i-1]-'0'+sstr[sstr.size()-i-1]-'0'+adder;
            } else {
                cond = lstr[lstr.size()-i-1]-'0'+adder;
            }
            if (cond >= 2) {
                adder = 1;
                ans.push_back(cond-2);
            } else {
                adder = 0;
                ans.push_back(cond);
            }
        }
        if (adder == 1) {
            ans.push_back(1);
        }
        string anstr;
        for (auto i = ans.rbegin(); i != ans.rend(); ++i) {
            anstr.push_back(char(int(*i)+'0'));
        }
        return anstr;
    }
};
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