Description of the problem:

The string `"PAYPALISHIRING"` 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   A   H   N
A P L S I I G
Y   I   R
```

And then read line by line: `"PAHNAPLSIIGYIR"`

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

`string convert(string s, int numRows);`

Example 1:

```Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"
```

Example 2:

```Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:

P     I    N
A   L S  I G
Y A   H R
P     I```

The idea of solving the problem:

This problem is a bit troublesome to find a position directly from a string. We can store the zigzag pattern in a two-dimensional array and read the character from it and add it to a new string.

Look for a position directly: see Grandyang

Code:

```#define UP 1
#define DOWN 0
class Solution {
public:
string convert(string s, int numRows) {
if(numRows == 1)
return s;
int idx = 0;
int row = 0;
int direction = 1;
vector<vector<char>> zigzag(numRows);
while(idx < s.size()){
zigzag[row].push_back(s[idx++]);
if(row == (numRows-1) || row == 0)
direction ^= 1;
if(direction == UP){
row--;
}else if(direction == DOWN){
row++;
}
}
string ret;
for(int i = 0; i < numRows; i++){
for(char c: zigzag[i]){
ret.push_back(c);
}
}
return ret;
}
};```

Find the position directly:

```class Solution {
public:
string convert(string s, int nRows) {
if (nRows <= 1) return s;
string res = "";
int size = 2 * nRows - 2;
for (int i = 0; i < nRows; ++i) {
for (int j = i; j < s.size(); j += size) {
res += s[j];
int tmp = j + size - 2 * i;
if (i != 0 && i != nRows - 1 && tmp < s.size()) res += s[tmp];
}
}
return res;
}
};```