# 6.ZigZag-Conversion

## 6. ZigZag Conversion

## 题目地址

<https://leetcode.com/problems/zigzag-conversion/>

## 题目描述

```
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
```

## 代码

### Approach #1 Sort by Row

```java
class Solution {
  public String convert(String s, int numRows) {
        if (numRows == 1)        return s;

    List<StringBuilder> rows = new ArrayList();
    for (int i = 0; i < Math.min(numRows, s.length()); i++) {
      rows.add(new StringBuilder());
    }

    int curRow = 0;
    boolean goingDown = false;

    for (char c: s.toCharArray()) {
      rows.get(curRow).append(c);
      if (curRow == 0 || curRow == numRows - 1) {
        goingDown = !goingDown;
      }
      curRow += goingDown ? 1 : -1;
    }

    StringBuilder ret = new StringBuilder();
    for (StringBuilder row : rows) {
      ret.append(row);
    }
    return ret.toString();
  }
}
```

### Approach #2 Visit By Row

```java
class Solution {
  public String convert(String s, int numRows) {
    if (numRows == 1)        return s;
    StringBuilder ret = new StringBuilder();
    int n = s.length();
    int cycleLen = 2 * numRows - 2;

    for (int i = 0; i < numRows; i++) {
      for (int j = 0; j + i < n; j += cycleLen) {
        ret.append(s.charAt(j + i));
        // inner row
        if (i != 0 && i != numRows - 1 && j + cycleLen - i < n) {
          ret.append(s.charAt(j + cycleLen - i));
        }
      }
    }

    return ret.toString();
  }
}
```


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