<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 {
    const vector<pair<int, int>> directions{{0, 1}, {1, -1}};
public:
    string convert(string s, int numRows) {
        if(numRows == 1) return s;
        vector<vector<char>> solution{};
        int direction = 0;
        int x = 0;
        int y = 0;
        for(char ch : s) {
            if(x >= solution.size()) {
                solution.push_back(vector<char>(numRows, '\0'));
            }
            solution[x][y] = ch;
            int shift = y + directions[direction].second;
            if(shift < 0 || shift >= numRows) {
                direction = direction != 0 ? 0 : 1;
            }
            x += directions[direction].first;
            y += directions[direction].second;
        }
        string result{};
        for(int j = 0; j < numRows; j++) {
            for(int i = 0; i < solution.size(); i++) {
                if(solution[i][j] != '\0') result += solution[i][j];
            }
        }
        return result;
    }
};
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