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0006.py
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0006.py
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# Source: https://leetcode.com/problems/zigzag-conversion
# Title: ZigZag Conversion
# Difficulty: Medium
# Author: Mu Yang <http://muyang.pro>
################################################################################################################################
# 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
#
# 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
#
################################################################################################################################
class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
rows = [''] * min(numRows, len(s))
n = 2*(numRows-1)
for i, c in enumerate(s):
j = i % n
j = j if j < numRows else n-j
rows[j] += c
return ''.join(rows)