6.Zigzag Conversion

tags: Medium,String

6. Zigzag Conversion

題目描述

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);

範例

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

Example 3:

Input: s = "A", numRows = 1
Output: "A"

Constraints:

• 1 <= s.length <= 1000
• s consists of English letters (lower-case and upper-case), ',' and '.'.
• 1 <= numRows <= 1000

解答

Python

class Solution:
    def convert(self, s: str, numRows: int) -> str:
        if numRows == 1:
            return s
        buf = []
        for _ in range(numRows):
            buf.append( [ 0 for _ in range(len(s))] )
        row,col = 0, 0
        direction = 'down'
        for c in s:
            #print(row,col,direction)
            buf[row][col] = c
            if row == numRows-1:
                direction = 'up'
                row, col = row-1, col+1
            elif row == 0:
                direction = 'down'
                row, col = row+1, col
            else:
                if direction == 'down':
                    row, col = row+1, col
                else:
                    row, col = row-1, col+1

        result = ''
        for i in range(numRows):
            for j in range(len(s)):
                #print(buf[i][j], end='')
                if buf[i][j] != 0:
                    result += buf[i][j]
            #print('')
        return result           

玉山

C++

class Solution {
public:
    string convert(string s, int numRows) {
        if (numRows == 1) return s;
        string ans;
        int step = (numRows - 1) * 2;
        for (int i = 0; i < numRows; i++) {
            for (int j = 0; j < s.size(); j += step) {
                int left = j + i;
                int right = j + step - i;
                if (left < s.size()) ans += s[left];
                if (right < s.size() && left < right && right < j + step) ans += s[right];
            }
        }
        return ans;
    }
};

Yen-Chi ChenFri, Feb 3, 2023

Reference

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