6_ZigZag_Conversion.py

#!/usr/bin/python
# -*- coding: utf-8 -*-
"""
Desc: 6 Z字形变换 (中等)
Author: wangluyu
Date: 2020/1/8
"""

class Solution:
    """
    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 s, int numRows);
    Example 1:
    Input: s = "PAYPALISHIRING", numRows = 3
    Output: "PAHNAPLSIIGYIR"
    Example 2:
    Input: s = "PAYPALISHIRING", numRows = 4
    Output: "PINALSIGYAHRPI"
    Explanation:
    P     I    N
    A   L S  I G
    Y A   H R
    P     I
    将一个给定字符串根据给定的行数，以从上往下、从左到右进行 Z 字形排列。
    比如输入字符串为 "LEETCODEISHIRING" 行数为 3 时，排列如下：
    L   C   I   R
    E T O E S I I G
    E   D   H   N
    之后，你的输出需要从左往右逐行读取，产生出一个新的字符串，比如："LCIRETOESIIGEDHN"。
    请你实现这个将字符串进行指定行数变换的函数：
    string convert(string s, int numRows);
    示例 1:
    输入: s = "LEETCODEISHIRING", numRows = 3
    输出: "LCIRETOESIIGEDHN"
    示例 2:
    输入: s = "LEETCODEISHIRING", numRows = 4
    输出: "LDREOEIIECIHNTSG"
    解释:
    L     D     R
    E   O E   I I
    E C   I H   N
    T     S     G
    """
    def convert(self, s: str, numRows: int) -> str:
        # TODO Z字形变换
        pass