主要是一个Z字形状

0    10

1   911

2  8 12

3 7  13

46   14

5     15。。。。。这样一个形状

因为只需要横向顺序，所以横向压缩就好

上面变成这样

0，  ，10

1，9，11

2，8，12

3，7，13

4，6，14

5，  ，15

这样的话只需要把数组纵向存储然后空出奇数行首位并注意顺序即可

c# code

public class Solution
{
    public string Convert(string s, int numRows)
    {
        if (numRows == 1 || s.Length <= numRows)
            {
                return s;
            }
            string ret = “”;
            if (numRows == 2)
            {
                for (int i = 0; i < s.Length && 2 * i < s.Length; i++)
                {
                    ret += s[2 * i];
                }
                for (int i = 0; i < s.Length && 2 * i + 1 < s.Length; i++)
                {
                    ret += s[2 * i + 1];
                }
            }
            else
            {
                char[,] c = new char[numRows, //s.Length % (numRows * 2 – 2) == 0 ? s.Length / (numRows * 2 – 2) :
                    (s.Length / (numRows * 2 – 2) + 1) * 2];
                int q = numRows < 3 ? 1 : numRows – 1;
                int index = 0, row = 0, col = 0;
                while (index < s.Length)
                {
                    if (col % 2 == 0)
                    {
                        for (row = 0; row < numRows && index < s.Length; row++)
                        {
                            c[row, col] = s[index];
                            index++;
                        }
                        col++;
                    }
                    else
                    {
                        for (row = numRows – 2; row > 0 && index < s.Length; row–)
                        {
                            c[row, col] = s[index];
                            index++;
                        }
                        col++;
                    }
                }
                for (int i = 0; i < c.GetLength(0); i++)
                {
                    for (int j = 0; j < c.GetLength(1); j++)
                    {
                        if (c[i, j] != ‘\0’)
                        {
                            ret += c[i, j];
                        }
                    }
                }
            }
            return ret;
    }
}