[LeetCode] ZigZag Conversion

Problem

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 text, int nRows);

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

Note

很奇葩的题目，PAYPAL IS HIRING.
先将s转化为字符数组c，便于操作。
再看nRows要求转化为多少行，就建立多大的StringBuilder数组sb[]。注意，建立完之后要用for循环对每一个StringBuilder[i]初始化：sb[i] = new StringBuilder();
然后用指针i循环字符数组c：顺序从0到nRows-1每一行放入一个字符c[i++]，然后逆序从nRows-2到1每一行放入一个字符c[i++]，直到i == c.length.
以上就将竖向ZigZag形式的字符串以横向读取顺序存入了数组sb中，将sb[i]中的所有字符串都合并到sb[0]中即可。

Solution

public class Solution {
    public String convert(String s, int nRows) {
        char[] c = s.toCharArray();
        int len = c.length;
        StringBuffer[] sb = new StringBuffer[nRows];
        for (int i = 0; i < sb.length; i++) sb[i] = new StringBuffer();
    
        int i = 0;
        while (i < len) {
            for (int idx = 0; idx < nRows && i < len; idx++) // vertically down
                sb[idx].append(c[i++]);
            for (int idx = nRows-2; idx >= 1 && i < len; idx--) // obliquely up
                sb[idx].append(c[i++]);
        }
        for (int idx = 1; idx < sb.length; idx++)
            sb[0].append(sb[idx]);
        return sb[0].toString();
    }
}