【leetcode】Z字形变换【python】

解题思路，按照列来存放，按照行来遍历就行。列的步长根据行变化，在数据结构上不需要设计成z字形的数组
坑1：二维数组创建错误：具体参考上一篇博客
坑2：numRows=1的情况下，步长设置错误

            if numRows == 1:
                return s

AC解答

class Solution:
    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """
        rows = [(['!'] * 100) for i in range (numRows)]
        rowindex = [0] * numRows #记录每一行元素的个数
        row = 0
        for i in s:
            if row == 0:
                step = 1
            if row == numRows - 1:
                step = -1
            if numRows == 1:
                return s
            rows[row][rowindex[row]] = i
            rowindex[row] += 1
            row += step
        #print(rows)
        str1 = ''
        for i in range(numRows):
            for j in rows[i]:
                if j != '!':
                    str1 = str1 + j
        return str1