6. ZigZag Conversion

题目：

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".

代码：

看了半天，都没怎看懂题目。查了一下，原来：

Zigzag:即循环对角线结构（

0				8				16
1			7	9			15	17
2		6		10		14		18
3	5			11	13			19
4				12				20

）

这样子的话，就不算麻烦了，只要正确的字符串操作，把每个元素加在正确的位置就好。

于是，共有numRows行，所以定义一个保护numRows元素的数组，用来保存每一行的字符串。

观察规律，第一列从0写到numRows-1，之后从后往前，及numRows-2到0开始逐渐添加。输出并不需要考虑宽度，所以对字符串来说，之后往后加就好

//凑了半天，总算凑出来了

def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """
        i = 0
        j = 0
        gap = numRows-2
        temp= ['' for x in range(numRows)]
        while i<len(s):
            while(j<numRows):
                if(i>=len(s)):break
                temp[j] += str(s[i])
                j += 1
                i += 1
            j=gap
            while(j>0):
                if(i>=len(s)):break
                temp[j] += str(s[i])
                j -= 1
                i += 1
        result = ''
        for y in temp:result += y

return result

//网上查的，好像更直观   
    def convert2(self, s, nRows):
        if nRows==1: return s
        tmp=['' for i in range(nRows)]
        index=-1; step=1
        for i in range(len(s)):
            index+=step
            if index==nRows:
                index-=2; step=-1
            elif index==-1:
                index=1; step=1
            tmp[index]+=str(s[i])
        return ''.join(tmp)

测试了多次，两个速度差不多，必定时间复杂度和空间复杂度都一样：

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