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006_ZigZag_Conversion.py
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83 lines (70 loc) · 2.09 KB
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"""
6. ZigZag Conversion
Medium
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
"""
"""
first work through to see the pattern and get a better understanding:
s = "PAYPALISHIRING" numRows = 3
iteration 1 "P", row [0]
iteration 2 "A", row [1] +1
iteration 3 "Y", row [2] +1
iteration 4 "P", row [1] -1
iteration 5 "A", row [0] -1
iteration 6 "L", row [1] +1
iteration 7 "I", row [2] +1
...
--> row == 0 : step = 1
row == numRows -1 (hit the end row): step -1
row = row + step
"""
"""
EX: s = "PAY"
when numRows == 1 : return s : "PAY"
when numRows >=1 lem(s): return s:
"P
A
Y"
"""
class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1 or numRows >= len(s):
return s
# create a list of empty strings
zigzag = ['' for x in range(numRows)]
# initialize the sequence. row = row index
row, step = 0 , 1
## add s[character] to the list using row index##
for char in s:
# add character to it's row in zigzag
zigzag[row] += char
if row == 0:
step = 1
elif row == numRows -1:
step = -1
row += step
#join the list together. this enable us to return the result the way we want
return ''.join(zigzag)
"""
Success
Details
Runtime: 52 ms, faster than 98.32% of Python3 online submissions for ZigZag Conversion.
Memory Usage: 13.8 MB, less than 9.13% of Python3 online submissions for ZigZag Conversion.
"""