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 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
Solution:–
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++;
currStringIndex++;
}
currRow -= 2;
currCol++;
// Move up (with moving right also).
while (currRow > 0 && currCol < numCols && currStringIndex < n) {
matrix[currRow][currCol] = s.charAt(currStringIndex);
currRow--;
currCol++;
currStringIndex++;
}
}
StringBuilder answer = new StringBuilder();
for (char[] row: matrix) {
for (char character: row) {
if (character != ' ') {
answer.append(character);
}
}
}
return answer.toString();
}
}
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