6. Z 字形变换_ZigZag Conversion

ZigZag Conversion

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。

题目来源：力扣（LeetCode）链接：https://leetcode-cn.com/problems/zigzag-conversion

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

python3:

class Solution:
    def convert(self, s: str, numRows: int) -> str:
        str_len = len(s)
        convert_list = [ [''] * str_len for i in range(numRows)]
        i = 0
        j = 0
        count = 0
        up_to_down = True
        while count < str_len:
            convert_list[j][i] = s[count]
            count = count + 1
            if up_to_down:
                j = j + 1
            else:
                j = j - 1
                i = i + 1
            
            if j >= numRows:
                i = i + 1
                j = j - 2
                up_to_down = False
            elif j < 0:
                j = j + 2
                up_to_down = True
        res_str = ''
        for lists in convert_list:
            for l in lists:
                res_str = res_str + l
        return res_str

改进：

class Solution:
    def convert(self, s: str, numRows: int) -> str:
        result_list = [''] * numRows
        i = 0
        j = 0
        count = 0
        up_to_down = True
        str_len = len(s)
        while count < str_len:
            result_list[j] = result_list[j] + s[count]
            count = count + 1
            if up_to_down:
                j = j + 1
            else:
                j = j - 1
                i = i + 1
            
            if j >= numRows:
                i = i + 1
                j = j - 2
                up_to_down = False
            elif j < 0:
                j = j + 2
                up_to_down = True
        res_str = ''
        for res in result_list:
            res_str = res_str + res
        return res_str

标签：count,down,Conversion,字形,res,up,ZigZag,numRows,str
来源： https://blog.csdn.net/qq_15706279/article/details/100585359