Z transformation
Arrange a given string s in a zigzagging order from top to bottom and left to right, numRows, according to the given number of rows.
For example, if the string “PAYPALISHIRING” is set to 3 rows, it will look like this:
PAHNAPLSIIGYIR Then your output needs to be read line by line from left to right, producing A new string, such as “PAHNAPLSIIGYIR”.
Implement this function to convert a string to the specified number of lines:
string convert(string s, int numRows);
Examples can be found on the LeetCode website.
Source: LeetCode link: leetcode-cn.com/problems/zi… Copyright belongs to the Collar buckle network. Commercial reprint please contact official authorization, non-commercial reprint please indicate the source.
Solution 1: traversal
The first step is to determine if the string length is empty or less than 3, or if numRows is 1 (that is, all tiled on one line). Otherwise, iterate over the string’s characters and mark them with a two-dimensional Boolean array in 2 steps. First, go down and no more, then slash to the right. Repeat these 2 steps until all the characters are recorded, and return the result based on the Boolean array’s mark.
public class Solution {
public static String convert(String s, int numRows) {
if (s == null || s.length() < 3 || numRows == 1) {
return s;
}
char[][] chars = new char[numRows][s.length()];
boolean[][] flags = new boolean[numRows][s.length()];
int x = -1, y = 0, count = 0;
while (count < s.length()) {
while (x < numRows && count < s.length()) {
if (x + 1 < numRows) {
x++;
chars[x][y] = s.charAt(count);
flags[x][y] = true;
count++;
} else {
break; }}while (x >= 0 && count < s.length()) {
if (x - 1> =0) {
x--;
y++;
chars[x][y] = s.charAt(count);
flags[x][y] = true;
count++;
} else {
break;
}
}
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < chars.length; i++) {
for (int j = 0; j < chars[i].length; j++) {
if(flags[i][j]) { sb.append(String.valueOf(chars[i][j])); }}}return sb.toString();
}
public static void main(String[] args) {
System.out.println(convert("ABC".1)); }}Copy the code