# 6. Zigzag Conversion

## Description

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

## Solutions

Solution 1: Simulation

We use a two-dimensional array $g$ to simulate the process of the $Z$-shape arrangement, where $g[i][j]$ represents the character at the $i$-th row and the $j$-th column. Initially, $i=0$, and we define a direction variable $k$, initially $k=-1$, indicating moving upwards.

We traverse the string $s$ from left to right. Each time we traverse to a character $c$, we append it to $g[i]$. If $i=0$ or $i=numRows-1$ at this time, it means that the current character is at the turning point of the $Z$-shape arrangement, and we reverse the value of $k$, i.e., $k=-k$. Next, we update the value of $i$ to $i+k$, i.e., move up or down one row. Continue to traverse the next character until we have traversed the string $s$, and we return the string concatenated by all rows in $g$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.

• 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();
}
}

• class Solution {
public:
string convert(string s, int numRows) {
if (numRows == 1) return s;
string ans;
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.push_back(s[idx]);
idx += interval;
interval = group - interval;
if (interval == 0) interval = group;
}
}
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)

• 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;
}
const g = new Array(numRows).fill(_).map(() => []);
let i = 0;
let k = -1;
for (const c of s) {
g[i].push(c);
if (i === 0 || i === numRows - 1) {
k = -k;
}
i += k;
}
return g.flat().join('');
};

• public class Solution {
public string Convert(string s, int numRows) {
if (numRows == 1) {
return s;
}
int n = s.Length;
StringBuilder[] g = new StringBuilder[numRows];
for (int j = 0; j < numRows; ++j) {
g[j] = new StringBuilder();
}
int i = 0, k = -1;
foreach (char c in s.ToCharArray()) {
g[i].Append(c);
if (i == 0 || i == numRows - 1) {
k = -k;
}
i += k;
}
StringBuilder ans = new StringBuilder();
foreach (StringBuilder t in g) {
ans.Append(t);
}
return ans.ToString();
}
}

• 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
}
}

• class Solution {
/**
* @param string $s * @param int$numRows
* @return string
*/

function convert($s,$numRows) {
if ($numRows == 1 || strlen($s) <= $numRows) { return$s;
}

$result = '';$cycleLength = 2 * $numRows - 2;$n = strlen($s); for ($i = 0; $i <$numRows; $i++) { for ($j = 0; $j +$i < $n;$j += $cycleLength) {$result .= $s[$j + $i]; if ($i != 0 && $i !=$numRows - 1 && $j +$cycleLength - $i <$n) {
$result .=$s[$j +$cycleLength - $i]; } } } return$result;
}
}