<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 numRows) {
        vector<pair<char, int>> dp(s.size());
        int i=0, currRow=1;

        bool up = false, start=true;
        if(numRows==1) return s;
        while(i<s.size()){
            dp[i].first = s[i];
            dp[i].second = currRow;

            if(currRow==numRows){
                currRow--;
                up = !up;
            }
            else if(currRow==1){
                currRow++;
                if(!start)
                    up = !up;
                else
                    start = false;
            }
            else if(up){
                currRow--;
            }
            else currRow++;

            i++;
        }

        string res = "";

        for(int x=1; x<=numRows; x++){
            for(int i=0; i<dp.size(); i++){
                if(dp[i].second == x) res += dp[i].first;
            }
        }

        return res;
    }
};
12181