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 <= 1000s consists of English letters (lower-case and upper-case), ',' and '.'.1 <= numRows <= 1000
Solutions
Solution 1: Simulation
We use a 2D array $g$ to simulate the process of arranging the string in a zigzag pattern, where $g[i][j]$ represents the character at row $i$ and column $j$. Initially, $i = 0$. We also define a direction variable $k$, initially $k = -1$, which means moving upwards.
We traverse the string $s$ from left to right. For each character $c$, we append it to $g[i]$. If $i = 0$ or $i = \textit{numRows} - 1$, it means the current character is at a turning point in the zigzag pattern, so we reverse the value of $k$, i.e., $k = -k$. Then, we update $i$ to $i + k$, which means moving up or down one row. Continue traversing the next character until the end of the string $s$. Finally, we return the concatenation of all rows in $g$ as the result.
The time complexity is $O(n)$ and the space complexity is $O(n)$, where $n$ is the length of the string $s$.
PythonJavaC++GoTypeScriptRustJavaScriptC#CPHP
class Solution:
def convert(self, s: str, numRows: int) -> str:
if numRows == 1:
return s
g = [[] for _ in range(numRows)]
i, k = 0, -1
for c in s:
g[i].append(c)
if i == 0 or i == numRows - 1:
k = -k
i += k
return ''.join(chain(*g))(code-box)
class Solution {
public String convert(String s, int numRows) {
if (numRows == 1) {
return s;
}
StringBuilder[] g = new StringBuilder[numRows];
Arrays.setAll(g, k -> new StringBuilder());
int i = 0, k = -1;
for (char c : s.toCharArray()) {
g[i].append(c);
if (i == 0 || i == numRows - 1) {
k = -k;
}
i += k;
}
return String.join("", g);
}
}(code-box)
class Solution {
public:
string convert(string s, int numRows) {
if (numRows == 1) {
return s;
}
vector<string> g(numRows);
int i = 0, k = -1;
for (char c : s) {
g[i] += c;
if (i == 0 || i == numRows - 1) {
k = -k;
}
i += k;
}
string ans;
for (auto& t : g) {
ans += t;
}
return ans;
}
};(code-box)
func convert(s string, numRows int) string {
if numRows == 1 {
return s
}
g := make([][]byte, numRows)
i, k := 0, -1
for _, c := range s {
g[i] = append(g[i], byte(c))
if i == 0 || i == numRows-1 {
k = -k
}
i += k
}
return string(bytes.Join(g, nil))
}(code-box)
function convert(s: string, numRows: number): string {
if (numRows === 1) {
return s;
}
const g: string[][] = new Array(numRows).fill(0).map(() => []);
let i = 0;
let k = -1;
for (const c of s) {
g[i].push(c);
if (i === numRows - 1 || i === 0) {
k = -k;
}
i += k;
}
return g.flat().join('');
}(code-box)
impl Solution {
pub fn convert(s: String, num_rows: i32) -> String {
if num_rows == 1 {
return s;
}
let num_rows = num_rows as usize;
let mut g = vec![String::new(); num_rows];
let mut i = 0;
let mut k = -1;
for c in s.chars() {
g[i].push(c);
if i == 0 || i == num_rows - 1 {
k = -k;
}
i = (i as isize + k) as usize;
}
g.concat()
}
}(code-box)
/**
* @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('');
};(code-box)
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();
}
}(code-box)
char* convert(char* s, int numRows) {
if (numRows == 1) {
return strdup(s);
}
int len = strlen(s);
char** g = (char**) malloc(numRows * sizeof(char*));
int* idx = (int*) malloc(numRows * sizeof(int));
for (int i = 0; i < numRows; ++i) {
g[i] = (char*) malloc((len + 1) * sizeof(char));
idx[i] = 0;
}
int i = 0, k = -1;
for (int p = 0; p < len; ++p) {
g[i][idx[i]++] = s[p];
if (i == 0 || i == numRows - 1) {
k = -k;
}
i += k;
}
char* ans = (char*) malloc((len + 1) * sizeof(char));
int pos = 0;
for (int r = 0; r < numRows; ++r) {
for (int j = 0; j < idx[r]; ++j) {
ans[pos++] = g[r][j];
}
free(g[r]);
}
ans[pos] = '\0';
free(g);
free(idx);
return ans;
}(code-box)
class Solution {
/**
* @param String $s
* @param Integer $numRows
* @return String
*/
function convert($s, $numRows) {
if ($numRows == 1) {
return $s;
}
$g = array_fill(0, $numRows, '');
$i = 0;
$k = -1;
$length = strlen($s);
for ($j = 0; $j < $length; $j++) {
$c = $s[$j];
$g[$i] .= $c;
if ($i == 0 || $i == $numRows - 1) {
$k = -$k;
}
$i += $k;
}
return implode('', $g);
}
}(code-box)
