> For the complete documentation index, see [llms.txt](https://jimmylin1991.gitbook.io/practice-of-algorithm-problems/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://jimmylin1991.gitbook.io/practice-of-algorithm-problems/string/6.-zigzag-conversion.md).

# 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
```

```cpp
string convert(string s, int numRows) { // time: O(n); space: O(n)
    if (s.empty()) return "";
    int n = s.length(), i = 0;
    vector<string> tmp(numRows, "");
    while (i < n) {
        // Vertical
        for (int j = 0; j < numRows && i < n; ++j) {
            tmp[j].push_back(s[i++]);
        }
        // Oblique
        for (int j = numRows - 2; j >= 1 && i < n; --j) {
            tmp[j].push_back(s[i++]);
        }
    }
    string res;
    for (const string& str : tmp) res += str;
    return res;
}
```

```cpp
string convert(string s, int numRows) { // time: O(n); space: O(n)
    if (s.empty() || numRows <= 1) return s;
    int n = s.length(), size = 2 * numRows - 2;
    string res;
    for (int i = 0; i < numRows; ++i) {
        for (int j = i; j < n; j += size) {
            res += s[j];
            int pos = j + size - 2 * i;
            if (i != 0 && i != numRows - 1 && pos < n) res += s[pos];
        }
    }
    return res;
}
```


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