題目（來自leetcode網站）：

實現了python版本

實現了C++版本

題目含義為，將輸入字串 首先根據numRows大小進行 列的 Z 字形 排列 後，把每一行直接拼接起來輸出；

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

解答：

#我的解法比較low 是根據人思考的邏輯來的，希望高手大佬指點一二
#python版本
class Solution:
    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """
        a=[]

        if len(s)<=numRows or numRows==1:
            return s
        for i in range(numRows):
            a.append([])
        i=0
        j=0
        while(i<len(s)):
            for k in range(numRows):
                if j == 0:
                    a[k].append(s[i])
                    i+=1
                    if i >=len(s):
                        break
                else:
                    a[numRows-j-1].append(s[i])
                    i+=1
                    break
            j+=1
            if j % (numRows-1)==0:
                j = 0
        b=""
        for i in range(numRows):
            print(a[i])
            for j in range(len(a[i])):
                b+=a[i][j]
        return b
                    
                
 
#C++版本
class Solution {
public:
    string convert(string s, int numRows) {
        //string** a=new string*[numRows];
        string a[numRows];
        if(s.size()<=numRows || numRows ==1){
            return s;
        }

        
        int i=0;
        int j=0;
        cout<<a[0]<<endl;
        while(i < s.size()){
            for(int k=0;k<numRows;k++){
                if(j==0){
                    a[k]+=s[i];
                    i++;
                    if(i >=s.size()){
                        break;
                    }
                }else{
                    a[numRows-j-1]+=s[i];
                    i++;
                    break;
                }
            }
            j++;
            if(j%(numRows-1)==0){
                j=0;
            }
        }
        string out;
        for(int i=0;i<numRows;i++){
            out+=a[i];
            }
        
        return out;
        
    }
};