<p>The string <code>&quot;PAYPALISHIRING&quot;</code> is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)</p>

<pre>
P   A   H   N
A P L S I I G
Y   I   R
</pre>

<p>And then read line by line: <code>&quot;PAHNAPLSIIGYIR&quot;</code></p>

<p>Write the code that will take a string and make this conversion given a number of rows:</p>

<pre>
string convert(string s, int numRows);
</pre>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;PAYPALISHIRING&quot;, numRows = 3
<strong>Output:</strong> &quot;PAHNAPLSIIGYIR&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;PAYPALISHIRING&quot;, numRows = 4
<strong>Output:</strong> &quot;PINALSIGYAHRPI&quot;
<strong>Explanation:</strong>
P     I    N
A   L S  I G
Y A   H R
P     I
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;A&quot;, numRows = 1
<strong>Output:</strong> &quot;A&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s</code> consists of English letters (lower-case and upper-case), <code>&#39;,&#39;</code> and <code>&#39;.&#39;</code>.</li>
	<li><code>1 &lt;= numRows &lt;= 1000</code></li>
</ul>

class Solution {
public:
    string convert(string s, int n) {
        if (n == 1) {
            return s;
        }

        // first, last row = n - 2 = 2
        // 2nd = n - 3 = 1, n - 2 - (n - 3) - 1 = 0
        // 3nd = 

        // 1 = 3, 0 = n - 1, n - (n - 1) - 1
        // 2 = 2, 1 = prev - 1, n - (n - 2) - 1
        // 3 = 1, 2 = prev - 1, n - (n - 3) - 1
        // 4 = 0, 3

        ostringstream oss;
        int g1 = n - 1;
        int g2 = 0;

        for (int i = 0; i < n; i++) {
            bool useG1 = true;

            for (int j = i; j < s.size();) {
                oss << s[j];

                if (useG1 && g1 == 0) useG1 = false;
                else if (!useG1 && g2 == 0) useG1 = true;

                if (useG1) {
                    j += g1 * 2;
                } else {
                    j += g2 * 2;
                }

                useG1 = !useG1;
            }

cout << oss.str() << endl;

            g1 -= 1;
            g2 += 1;
        }

        return oss.str();
    }
};
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