# 6. 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 s, int numRows);

Example 1:

Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:

P     I    N
A   L S  I G
Y A   H R
P     I

# Solution

Approach 1: Sort by Row.

Approach 2: Visit by row.

# Code (Python)

Approach 1:

Approach 2:

# Code (C++)

Approach 1:

// Sort by Row.
class Solution {
public:
    string convert(string s, int numRows) {
        string row[numRows];
        for (int i = 0; i < numRows; ++i)
            row[i] = "";
/*
        int i = 0;
        while (i < s.size())
        {
            for (int r = 0; r < numRows && i < s.size(); ++r, ++i)
            {
                row[r] += s[i];
            }
            for (int r = numRows - 2; r > 0 && i < s.size(); --r, ++i)
            {
                row[r] += s[i];
            }
        }
*/
        if (numRows == 1) return s; // Need to handle the case of "numRows == 1" separately.
        int step = -1; // Initialize it as -1.
        int r = 0;
        for (int i = 0; i < s.size(); ++i)
        {
            row[r] += s[i];
            if (r == numRows - 1 || r == 0)
                step = 0 - step;
            r += step;
        }

        string res = "";
        for (int i = 0; i < numRows; ++i)
            res += row[i];
        return res;
    }
};

Approach 2:

// Visit by row.
// For example:
// 0     6       12   ==  0     6     12
// 1   5 7    11          1  5  7  11
// 2 4   8 10             2  4  8  10
// 3     9                   3      9
// Characters in row 0 are located at indexes 2 * (numRows - 1) * col.
// Characters in inner row are located at indexes 2 * (numRows - 1) * col +- row.
// Characters in row (numRows - 1) are located at indexes 2 * (numRows - 1) * col + (numRows - 1).
class Solution {
public:
    string convert(string s, int numRows) {
        if (numRows == 1) return s; // Need to handle the case of "numRows == 1" separately.
        string res = "";
        for (int col = 0; ; ++col)
        {
            int idx = 2 * (numRows - 1) * col;
            if (idx >= s.size())
                break;
            res += s[idx];
        }
        for (int row = 1; row < numRows - 1; ++row)
        {
            for (int col = 0; ; ++col)
            {
                int idx1 = 2 * (numRows - 1) * col - row;
                int idx2 = 2 * (numRows - 1) * col + row;
                if (idx1 >= s.size() && idx2 >= s.size())
                    break;
                if (idx1 >= 0)
                    res += s[idx1];
                if (idx2 < s.size())
                    res += s[idx2];
            }
        }
        for (int col = 0; ; ++col)
        {
            int idx = 2 * (numRows - 1) * col + (numRows - 1);
            if (idx >= s.size())
                break;
            res += s[idx];
        }
        return res;
    }
};