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".

分析：

　　這道題主要是找規律。

方法一：

　　比較直觀的解法，使用list存儲每一行，最後拼接。

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

方法二：

　　發現所有行的重復周期都是 2 * nRows - 2，對於首行和末行之間的行，還會額外重復一次，重復的這一次距離本周期起始字符的距離是 2 * nRows - 2 - 2 * i

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

ZigZag Conversion