> For the complete documentation index, see [llms.txt](https://a.b.cr/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://a.b.cr/dictionary/algorithm/diary/2022/06/2022-06-18.md).

# 2022-06-18

## [6. Zigzag Conversion](https://leetcode.com/problems/zigzag-conversion/)

### Description

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

#### Approach #0

```go
func convert(s string, numRows int) string {
    n, r := len(s), numRows
    if r == 1 || n <= r {
        return s
    }
    t := 2*r - 2
    c := int(math.Ceil(float64(n)/float64(t))) * (r - 1)
    mat := make([][]byte, r)
    for i := 0; i < r; i++ {
        mat[i] = make([]byte, c)
    }
    x, y := 0, 0
    for i, ch := range s {
        mat[x][y] = byte(ch)
        if i%t < r-1 {
            x++
        } else {
            x--
            y++
        }
    }
    var ans []byte
    for i := 0; i < r; i++ {
        for j := 0; j < c; j++ {
            if mat[i][j] > 0 {
                ans = append(ans, mat[i][j])
            }
        }
    }
    return string(ans)
}
```


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