[leetcode] No 6. Zigzag Conversion Python

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[leetcode] No 6. Zigzag Conversion Python

for dream 2023. 2. 20. 22:27

[Problem]

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)

[Explanation]

After arranging string to zigzag, combine zigzagged string as one.

There is a given number of rows. So I can arrange string in codition of rows and combine as one

For example, 'PAYPALISHIRING' and a given rows is 3

row 1. P A H N > (rearrange) > PAHN

row 2. A P L S I I G > > APLSIIG

row 3. Y I R > > YIR

>>> (combine) >>>> PAHNAPLSIIGYIR (It is correct answer!)

[Solution]

1. First, string has to rearrange as zigzag. There are directions such as up and down.

So, I consider direction as defined 'd'.

2. In each row, an arranged character has to be combined with a previous character.

In this part, I have to consider the location of row, 'rows'.

If the initial row allocation is empty, 'append' is used.

If not, I just add a new character to the previous string located to the specifc row.

3. If a direction is down, 'rows' is increased by 1.

On the other hands, a direction is up, 'rows' is decreased by 1.

If 'rows' is approched to the limited number like 0 or a number of row, it has to be controled as limited range.

Like, 'rows = min(max(0, rows), numRows-1)'

4. Finally, all the rearranged string which are in each row have to be combined.

class Solution:
    def convert(self, s: str, numRows: int) -> str:
        word = []
        rows = 0
        d = 1

        for i in range(len(s)):
            if len(word) < numRows:
                word.append(s[i])
            else:
                word[rows] = word[rows] + s[i]

            if d == 1: rows += 1
            else: rows -= 1

            if rows == numRows:
                rows = numRows - 2
                d = 0
            elif rows == -1:
                rows = 1
                d = 1

            rows = min(max(0, rows), numRows-1)


        ans = ''

        for j in range(min(len(s),numRows)):
            ans = ans + word[j]

        return ans