> For the complete documentation index, see [llms.txt](https://zhenchaogan.gitbook.io/leetcode-solution/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://zhenchaogan.gitbook.io/leetcode-solution/leetcode-6-zigzag-conversion.md).

# LeetCode 6. ZigZag Conversion

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
```

**Example 3:**

```
Input: s = "A", numRows = 1
Output: "A"
```

**Constraints:**

* `1 <= s.length <= 1000`
* `s` consists of English letters (lower-case and upper-case), `','` and `'.'`.
* `1 <= numRows <= 1000`

## Solution:

[English Version in Youtube](https://youtu.be/1v4c3nj_8jc)

[中文版解答Youtube Link](https://youtu.be/eG9JVMAhKz8)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1iv4y1Z7M9/)

```
class Solution {
public:
    string convert(string s, int numRows) {
        if (numRows == 1) {
            return s;
        }
        
        vector<string> rows(numRows);
        int row = 0;
        int step = 1;
        for (char ch : s) {
            rows[row].push_back(ch);
            
            if (row == (numRows - 1)) {
                step = -1;
            } else if (row == 0) {
                step = 1;
            }
            row += step;
        }
        
        stringstream ss;
        for (const string& row : rows) {
            ss << row;
        }
        return ss.str();
    }
};
```