6 ZigZig Conversion[M]Z字形变换

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)

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：
  Input: s= "ABCDEFGHIJKLMNOP", numRows = 4,
  Output:AGMBFHLNCEIKODJP
  Explanation：
    A G M
    B F H L N
    C E I K O
    D J P

思路：

先将字符串\(s\)进行Z字排列，再按行读取字符。将Z字分为上中下三个部分，则例子的Z上：ABCD，中：EF，下:GHIJ。分析其规律可以发现，对于指定的行数\(n\)，中部的个数为\(n-2\)，将每一个上+中作为一个循环，则循环周期为\(t=n+n-2\)。于是有
第1行：\(s[0], s[0+t],s[0+2 \cdot t], \cdots ;\)
第2行：\(s[1], s[0+t-1],s[1+t],\cdots;\)
\(\cdots\)
第\(n\)行：\(s[n-1],s[n-1+t],s[n-1+2 \cdot t]\cdots .\)

C++

class solution{
public:
  string convert(string s, int numRows){
    if(s.size()==1) return s;
    string resString;
    int stringLen=s.size();
    int cycleLen=2*numRows-2;

    for(int i=0;i<numRows;i++){
      for(int j=0;j+i<stringLen;j+=cycleLen){
        resString += s[i+j];
        if(i !=0 && i != numRows-1 && j+cycleLen-i < stringLen){
         resString+=s[j + cycleLen - i];
      }
    }
    return resString;
  }
};

python

class Solution(object):
    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """
        
        if numRows == 1:
            return s
        
        rows = [""]*numRows
        
        cycle = 2*numRows - 2
        result = ""
        
        level = 0 #层数
        aux = -1
        for i in s:
            rows[level] += i
            if(level == 0 or level == numRows -1):
                aux *= -1
            level += aux
        
        for i in rows:
            result += i
                    
        return result