[LeetCode][Python]ZigZag Conversion
# -*- coding: utf8 -*-'''
__author__ = 'dabay.wang@gmail.com'
https://oj.leetcode.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 text, int nRows);
convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".
===Comments by Dabay===
用nRows个数组来接受每一个字符。
当nRows等于1时，直接返回s。
当nRows大于等于2时，定义nRows个数组，
2*nRows-2 为一个循环。从第一行到最后一行，再从倒数第二行回到第二行。
用模来确定位置，
当小于等于nRows-1的时候，直接放到模对应的数组里面
当大于nRows-1的时候，这里是数学题了：
mod-(nRows-1)为应该往上数的行数，最后一行的标号为nRow-1，所以这个行标应该是：(nRow-1)-(mod-(nRows-1))=2*nRows-mod-2
'''
class Solution:
# @return a string
def convert(self, s, nRows):
if nRows == 1:
return s
array2d = []
for n in range(0, nRows):
array2d.append([])
for i in range(0, len(s)):
mod = i % (2*nRows-2)
if mod <= nRows-1:
array2d.append(s)
else:
x = 2*nRows - mod -2
array2d.append(s)
val = ""
for n in range(0, nRows):
for v in array2d:
val = val + v
return val
def main():
s = Solution()
print s.convert("PAYPALISHIRING", 3)
if __name__ == "__main__":
import time
start = time.clock()
main()
print "%s sec" % (time.clock() - start)
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