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等差数列的首项和公差

    第一行为首项是0，公差是2（n-1）的等差数列
    中间是两个等差数列交错
    最后一行是首项是n-1,公差是2(n-1)的等差数列

class Solution:
    def convert(self, s: str, n: int) -> str:
        if n == 1: return s # n是输出的行数
        res = ''
        for i in range(n):
            if not i or i == n - 1: # 当前是第一行或者最后一行时
                j = i
                while j < len(s): 
                    res += s[j]
                    j += 2 * (n - 1)
            else: # 当前行是中间行，有两个交错的等差数列
                j = i; k = 2 * (n - 1) - i  # j,k分别是前后等差数列的值
                while j < len(s) or k < len(s):
                    if j < len(s): res += s[j]
                    if k < len(s): res += s[k]
                    j += 2 * (n - 1)
                    k += 2 * (n - 1)
        return res