ZigZag Conversion Solutions in C++

// Source : https://oj.leetcode.com/problems/zigzag-conversion/
// Author : Hao Chen
// Date   : 2014-07-17



#include <iostream>
#include <vector>
#include <string>
using namespace std;

string convert(string s, int nRows) {
    //The cases no need to do anything
    if (nRows<=1 || nRows>=s.size()) return s;
     
    vector<string> r(nRows);
    int row = 0;
    int step = 1;
    for(int i=0; i<s.size(); i ++) {
        if (row == nRows-1) step = -1;
        if (row == 0) step = 1;
        //cout << row <<endl;
        r[row] += s[i];
        row += step;
    }
    
    string result;
    for (int i=0; i<nRows; i++){
        result += r[i];
    }
    return result;
}

int main(int argc, char**argv){

    string s;
    int r;

    s = "PAYPALISHIRING";
    r = 3;
    cout << s << " : " << convert(s, 3) << endl;

}

#include <string>
using std::string;
#include <vector>
#include <numeric>


class Solution {
public:
    string convert(string s, int nRows) {
        if (s.empty() || nRows < 2) return s;
        std::vector<string> ret(nRows);
        for (size_t i=0; i<s.size(); ++i) {
            int m = i % (nRows-1), n = i / (nRows-1);
            (n & 0x1 ? ret[nRows-m-1] : ret[m]) += s[i];
        }
        return std::accumulate(ret.cbegin(), ret.cend(), string());
    }
};