[leetcode: Python]6. 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   NA P L S I I GY   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".

方法一：176ms
寻找字母下标与其所在行的对应关系：
编写一个函数convertRowNum(idx, numRows)

输入当前的字符下标idx与行数numRows，返回其所在的行号rowNum

函数如下：

def convertRowNum(self, idx, numRows):    return numRows -1 - abs(numRows - 1 - i % (2 * numRows - 2))

zigzag字符串中字母所在的行号序列为：

1, 2, ... , numRows - 1, numRows, numRows - 1, ... 2, 1

上述序列是一个长度为 2 * numRows - 2 的无限循环

class Solution(object):    def convert(self, s, numRows):        """        :type s: str        :type numRows: int        :rtype: str        """        if numRows == 1 or numRows >= len(s):            return s        final = [[] for row in xrange(numRows)]        for i in range(len(s)):            final[(numRows - 1 - abs(numRows - 1 - i % (2 * numRows - 2)))].append(s[i])        return ''.join([''.join(final[i]) for i in range(numRows)])

方法二：149ms

class Solution(object):    def convert(self, s, numRows):        """        :type s: str        :type numRows: int        :rtype: str        """        if numRows<=1:            return s        res = ""        n=len(s)        for i in range(numRows):            for j in range(i,n,2*(numRows-1)):                res += s[j]                if i>0 and i<numRows-1 and j+2*(numRows-1-i)<n:                    res +=s[j+2*(numRows-1-i)]        return res

方法三：116ms

class Solution(object):    def convert(self, s, numRows):        """        :type s: str        :type numRows: int        :rtype: str        """        if numRows <= 1 or len(s) <= numRows:            return s        i, j = 0, 1        ss = [''] * numRows        for v in s:            ss[i] += v            if i == numRows - 1:                j = -1            elif i == 0:                j = 1            i += j        return ''.join(ss)

方法四：92ms

class Solution(object):    def convert(self, s, numRows):        """        :type s: str        :type numRows: int        :rtype: str        """        if numRows == 1:            return s        i = 0        change = -1        newstr = ['']*numRows        for char in s:            if i == 0 or i==numRows-1:                change = -change            newstr[i] = newstr[i] + char            i = i + change        result = ''.join(newstr)        return result