【leetcode】Zigzag Conversion

Question ： 

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   NA P L S I I GY   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".

Anwser 1 ：     

class Solution { public:     string convert(string s, int nRows) {         // Start typing your C/C++ solution below         // DO NOT write int main() function             string ret(s);                  int stepArray[2];         int steps = 2 * nRows - 2;                  int retSize = 0;         for(int i = 0; i < nRows; i++)         {             int num = nRows - (i + 1) + nRows - (i + 1) - 1;             num++;             stepArray[0] = num;             if (stepArray[0] == 0)                 stepArray[0] = steps;             stepArray[1] = steps - num;             if (stepArray[1] == 0)                 stepArray[1] = steps;                              int j = i;             int index = 0;             while(j < s.size())             {                 ret[retSize++] = s[j];                 if (j == j + stepArray[index])                     break;                 j = j + stepArray[index];                 index = 1 - index;             }         }                      return ret;     } };

Anwser 2 ：     

class Solution {public:    string convert(string s, int nRows) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (nRows == 1) return s;                vector<vector<char>> v(nRows, vector<char>());                int i = 0;        int d = 0;                for (int j = 0; j < s.size(); j++) {            v[i].push_back(s[j]);                        if (d == 0) {                if (i + 1 == nRows) {                    i--;                    d = 1;                } else {                    i++;                }            } else {                if (i == 0) {                    i++;                    d = 0;                } else {                    i--;                }            }        }                string ret = "";                for (int j = 0; j < nRows; j++) {            for (int k = 0; k < v[j].size(); k++)                ret += v[j][k];        }                return ret;    }};

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