6. ZigZag Conversion

Problem:

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:

Squeeze the zigzag pattern horizontally to form a matrix. Now deal with the odd and even columns respectively.

For example let numRows be 5, if we list out the indecies:

row
 1    00    08    16
 2    01 07 09 15 17
 3    02 06 10 14 18
 4    03 05 11 13 19
 5    04    12    20

First calculate the matrix width:

pairs = floor( len(s) / (numRows + numRows - 2) )
width = pairs * 2 + ceil( (len(s) - pairs * (numRows + numRows - 2)) / numRows )

We can easily make a observation that the direction of odd and even columns and different.

Let the first column be index 0 and let i be the current position at column col.

We need to count the items between matrix[row][col] and matrix[row][col+1], exclusive.

next_i = i + (numRows - row) + (numRows - row), if col is even && 1 < row < numRows
next_i = i + row - 2 + row, if col is odd && 1 < row < numRows

If row == 1 or row == numRows, skip the odd columns.

next_i = i + numRows + (numRows - 2), if col is even && (row == 1 || row == numRows)
///**
 * @param {string} s
 * @param {number} numRows
 * @return {string}
 */let convert = function (s, numRows) {if (numRows <= 1) {return s;}const pairs = Math.floor(s.length / (numRows + numRows - 2));const width = pairs * 2 + Math.ceil((s.length - pairs * (numRows + numRows - 2)) / numRows);let result = '';for (let row = 1; row <= numRows; row++) {let i = row - 1;
        result += s[i] || '';for (let col = 0; col < width; col++) {if (row === 1 || row === numRows) {if (col % 2 === 0) {
                    i += numRows + (numRows - 2);} else {continue;}} else {if (col % 2 === 0) {
                    i += numRows - row + (numRows - row);} else {
                    i += row - 2 + row;}}
            result += s[i] || '';}}return result;};

☆: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:☆

☆: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:☆