# Question

Formatted question description: https://leetcode.ca/all/6.html

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


Example 3:

Input: s = "A", numRows = 1
Output: "A"


Constraints:

• 1 <= s.length <= 1000
• s consists of English letters (lower-case and upper-case), ',' and '.'.
• 1 <= numRows <= 1000

# Algorithm

Law: The first and last lines are added in the order of 2n-2. The position of the word diagonally across the line is the current column j+(2n-2)-2i (i is the index of the row).

# Code

• 
public class ZigZag_Conversion {

public class Solution {
public String convert(String s, int nRows) {
if (s == null || s.length() <= nRows || nRows == 1) {
return s;
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < nRows; i++) {
if (i == 0 || i == nRows - 1) {
int index = i;
while (index < s.length()) {
sb.append(s.charAt(index));
index += 2 * (nRows - 1);
}
} else {
int index = i;
while (index < s.length()) {
sb.append(s.charAt(index));
if (index + 2 * nRows - 2 * i - 2 < s.length()) {
sb.append(s.charAt(index + 2 * nRows - 2 * i - 2));
}
index += 2 * (nRows - 1);
}
}
}
return sb.toString();
}
}
}

############

class Solution {
public String convert(String s, int numRows) {
if (numRows == 1) {
return s;
}
StringBuilder ans = new StringBuilder();
int group = 2 * numRows - 2;
for (int i = 1; i <= numRows; i++) {
int interval = i == numRows ? group : 2 * numRows - 2 * i;
int idx = i - 1;
while (idx < s.length()) {
ans.append(s.charAt(idx));
idx += interval;
interval = group - interval;
if (interval == 0) {
interval = group;
}
}
}
return ans.toString();
}
}

• // OJ: https://leetcode.com/problems/zigzag-conversion/
// Time: O(N)
// Space: O(1)
class Solution {
public:
string convert(string s, int numRows) {
if (numRows == 1) return s;
int N = s.size(), d = (numRows - 1) * 2;
string ans;
for (int i = 0; i < numRows; ++i) {
int w = 2 * i;
for (int j = i; j < N;) {
ans += s[j];
w = d - w;
if (!w) w = d;
j += w;
}
}
return ans;
}
};

• class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
group = 2 * numRows - 2
ans = []
for i in range(1, numRows + 1):
interval = group if i == numRows else 2 * numRows - 2 * i
idx = i - 1
while idx < len(s):
ans.append(s[idx])
idx += interval
interval = group - interval
if interval == 0:
interval = group
return ''.join(ans)

############

class Solution(object):
def convert(self, s, numRows):
"""
:type s: str
:type numRows: int
:rtype: str
"""
if numRows <= 1:
return s
n = len(s)
ans = []
step = 2 * numRows - 2
for i in range(numRows):
one = i
two = -i
while one < n or two < n:
if 0 <= two < n and one != two and i != numRows - 1:
ans.append(s[two])
if one < n:
ans.append(s[one])
one += step
two += step
return "".join(ans)


• func convert(s string, numRows int) string {
if numRows == 1 {
return s
}
n := len(s)
ans := make([]byte, n)
step := 2*numRows - 2
count := 0
for i := 0; i < numRows; i++ {
for j := 0; j+i < n; j += step {
ans[count] = s[i+j]
count++
if i != 0 && i != numRows-1 && j+step-i < n {
ans[count] = s[j+step-i]
count++
}
}
}
return string(ans)
}

• function convert(s: string, numRows: number): string {
if (numRows === 1) {
return s;
}
const ss = new Array(numRows).fill('');
let i = 0;
let toDown = true;
for (const c of s) {
ss[i] += c;
if (toDown) {
i++;
} else {
i--;
}
if (i === 0 || i === numRows - 1) {
toDown = !toDown;
}
}
return ss.reduce((r, s) => r + s);
}


• /**
* @param {string} s
* @param {number} numRows
* @return {string}
*/
var convert = function (s, numRows) {
if (numRows == 1) return s;
let arr = new Array(numRows);
for (let i = 0; i < numRows; i++) arr[i] = [];
let mi = 0,
isDown = true;
for (const c of s) {
arr[mi].push(c);

if (mi >= numRows - 1) isDown = false;
else if (mi <= 0) isDown = true;

if (isDown) mi++;
else mi--;
}
let ans = [];
for (let item of arr) {
ans = ans.concat(item);
}
return ans.join('');
};


• using System.Collections.Generic;
using System.Linq;

public class Solution {
public string Convert(string s, int numRows) {
if (numRows == 1) return s;
if (numRows > s.Length) numRows = s.Length;
var rows = new List<char>[numRows];
var i = 0;
var j = 0;
var down = true;
while (i < s.Length)
{
if (rows[j] == null)
{
rows[j] = new List<char>();
}
j = j + (down ? 1 : -1);
if (j == numRows || j < 0)
{
down = !down;
j = j + (down ? 2 : -2);
}
++i;
}
return new string(rows.SelectMany(row => row).ToArray());
}
}

• impl Solution {
pub fn convert(s: String, num_rows: i32) -> String {
let num_rows = num_rows as usize;
if num_rows == 1 {
return s;
}
let mut ss = vec![String::new(); num_rows];
let mut i = 0;
let mut to_down = true;
for c in s.chars() {
ss[i].push(c);
if to_down {
i += 1;
} else {
i -= 1;
}
if i == 0 || i == num_rows - 1 {
to_down = !to_down;
}
}
let mut res = String::new();
for i in 0..num_rows {
res += &ss[i];
}
res
}
}