#6 Zigzag Conversion LeetCode Java Problem Solved | Java Leetcode Solution

Zigzag Conversion LeetCode Java Problem Solved | Java Leetcode Solution

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
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

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
P     I    N
A   L S  I G
Y A   H R
P     I

Example 3:

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


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


class Solution {
    public String convert(String s, int numRows) {
        if (numRows == 1) {
            return s;
        int n = s.length();
        int sections = (int) Math.ceil(n / (2 * numRows - 2.0));
        int numCols = sections * (numRows - 1);
        char[][] matrix = new char[numRows][numCols];
        for (char[] row: matrix) {
            Arrays.fill(row, ' ');
        int currRow = 0, currCol = 0;
        int currStringIndex = 0;
        // Iterate in zig-zag pattern on matrix and fill it with string characters.
        while (currStringIndex < n) {
            // Move down.
            while (currRow < numRows && currStringIndex < n) {
                matrix[currRow][currCol] = s.charAt(currStringIndex);
            currRow -= 2;
            // Move up (with moving right also).
            while (currRow > 0 && currCol < numCols && currStringIndex < n) {
                matrix[currRow][currCol] = s.charAt(currStringIndex);
        StringBuilder answer = new StringBuilder();
        for (char[] row: matrix) {
            for (char character: row) {
                if (character != ' ') {
        return answer.toString();

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