LeetCode-ZigZag Conversion

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".

class Solution {
public:
    string convert(string s, int nRows) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(s=="")return "";
        if(nRows<=1)return s;
        string ret;
        int round=2*nRows-2;//一个V字形有多少个字符 即样例中的P A Y P,全过程就是不断重复这个V字形
        int iter=s.length()/round;//多少格V字形
        int lastStart=iter*round;//最后一个V字形的开始
        //0
        for(int i=0;i<iter;i++){
            ret+=s[i*round];
        }
        if(lastStart<s.length())ret+=s[lastStart];
		//ret+="\n";
        for(int j=1;j<nRows-1;j++){
            for(int i=0;i<iter;i++){
                ret+=s[i*round+j];
                ret+=s[i*round+j+(nRows-1-j)*2];
            }
            if(lastStart+j<s.length())ret+=s[lastStart+j];
            if(lastStart+j+(nRows-1-j)*2<s.length())ret+=s[lastStart+j+(nRows-1-j)*2];
			//ret+="\n";
        }
        //last row
        for(int i=0;i<iter;i++){
            ret+=s[i*round+nRows-1];
        }
        if(lastStart+nRows-1<s.length())ret+=s[lastStart+nRows-1];
		//cout<<ret<<endl;
        return ret;
    }
};