Converting the Z-shaped Leetcode

Title Description

The number of a given string according to the given row, to down, left to right from the Z-shaped arrangement. Example, the input string is the number of rows "LEETCODEISHIRING" is 3, arranged as follows:

LCIR
ETOESIIG
EDHN
After that, you need the output from left to right read line by line, produce a new string, such as: "LCIRETOESIIGEDHN".

Thinking

1. Violent solution to simulate the transformation process in order to get an answer, but time-consuming, large memory footprint
2. Find the law, we find that the distance between the last line in the first two letters arranged to 2n-3, rather than the first law between the two letters is the last line: If the first character is defined as 0, the k-th character when the even-numbered, the next character pitch is 2 * (n- line-1), if k is odd, the next character is a distance 2 * line. This method only requires n times to traverse the string

Code

method one:

string convert(string s, int numRows) {
        int len = s.length();
        if (len<2 || numRows == 1)
            return s;
        string res;
        int numCols = len / (numRows + numRows - 2)*(numRows - 1);
        if (len % (numRows + numRows - 2) != 0)
        {
            if (len % (numRows + numRows - 2) <= numRows)
                numCols++;
            else
                numCols += 1 + len % (numRows + numRows - 2) - numRows;
        }
        vector<vector<char>> Matrix(numRows, vector<char>(numCols));
        int x = 0, y = 0;
        for (int i = 0; i<len; i++)
        {
            Matrix[x][y] = s[i];
            if (y % (numRows - 1) == 0)
            {
                if (x<numRows-1)
                    x++;
                else
                {
                    x--;
                    y++;
                }
            }
            else
            {
                y++;
                x--;
            }
        }
        for (int i = 0; i<numRows; i++)
        {
            for (int j = 0; j<numCols; j++)
            {
                if (Matrix[i][j] != '\0')
                    res += Matrix[i][j];
            }
        }
        return res;
    }

Method Two:

class Solution {
public:
    string convert(string s, int numRows) {
        int len = s.length();
        if (len<2 || numRows == 1)
            return s;
        string res;
        int line = 0;
        while (line<numRows)
        {
            int index = line;
            if (line == 0 || line == numRows - 1)
            {
                while (index<s.length())
                {
                    res += s[index];
                    index = index + 2 * (numRows-1) ;
                }
                line++;
            }
            else
            {
                int count = 0;
                while (index<s.length())
                {
                    res += s[index];
                    if (count % 2 == 0)
                        index = index + 2 * (numRows - line-1);
                    else
                        index = index + 2 * line;
                    count++;
                }
                line++;
            }
        }
        return res;
    }
};