## ZigZag Conversion

### 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:

```cpp
string convert(string text, int nRows);
```

`convert("PAYPALISHIRING", 3)` should return `"PAHNAPLSIIGYIR"`.

### 分析

n=4:

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

n=5:

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

### 代码

{% if book.java %}
```java
// ZigZag Conversion
// 时间复杂度O(n)，空间复杂度O(1)
public class Solution {
public String convert(String s, int numRows) {
if (numRows <= 1 || s.length() <= 1) return s;
StringBuilder result = new StringBuilder();
for (int i = 0; i < numRows; i++) {
for (int j = 0, index = i; index < s.length();
j++, index = (2 * numRows - 2) * j + i) {
result.append(s.charAt(index)); // 垂直元素
if (i == 0 || i == numRows - 1) continue; // 斜对角元素
if (index + (numRows - i - 1) * 2 < s.length())
result.append(s.charAt(index + (numRows - i - 1) * 2));
}
}
return result.toString();
}
}
```
{% endif %}

{% if book.cpp %}
```cpp
// LeetCode, ZigZag Conversion
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
string convert(string s, int nRows) {
if (nRows <= 1 || s.size() <= 1) return s;
string result;
for (int i = 0; i < nRows; i++) {
for (int j = 0, index = i; index < s.size();
j++, index = (2 * nRows - 2) * j + i) {
result.append(1, s[index]); // 垂直元素
if (i == 0 || i == nRows - 1) continue; // 斜对角元素
if (index + (nRows - i - 1) * 2 < s.size())
result.append(1, s[index + (nRows - i - 1) * 2]);
}
}
return result;
}
};
```
{% endif %}