ZigZag Convertion

#include<iostream>
#include<string>
#include<vector>
#include<math.h>
using namespace std;
 //从按行的角度出发，计算逐个需要存的字符的对应原串的下标。
class Solution {
public:Za
    string convert(string s, int numRows) {
string re= "";
int len =s.size();
int T=numRows*2-2;
double loops = len*1.0/T;
int idx =0;
if(numRows <=1||s.length()==0)
return s;


for(int i =0;i<numRows;i++)
for(int j =0;j<ceil(loops);j++)
if (i==0||i == numRows-1)
{
  idx= T*j+i;
  if (idx > len -1)
continue;
  re +=s[idx];
}
else
{
int idx1=T*j+i;
int idx2=T*j+i+2*(numRows-1-i);
// int idx2 = T*j+2*numRows-i-2;
   if(idx1>len -1)
continue;
re +=s[idx1];
if(idx2>len -1)
continue;
re +=s[idx2];
}

         return re;
        
    }
};




int main(){


string s ="PAYPALISHIRING";
Solution exp;
int row =3;
string re =exp.convert(s,row);
cout << re;
system("pause");
return 1;






}