day39 3208 zigzag scan (diagonal of matrix, find regularity)

3208. Zigzag Scan

In the image coding algorithm, a given square matrix needs to be $Z$ zigzag scanning (Zigzag Scan).

Given a $n\times$matrix of$n$ ,$The$ zigzag scanning process is shown in the figure below:

For the following$4\times$Matrix of$4$ ,

1 5 3 9
3 7 5 6
9 4 6 4
7 3 1 3

Its $The$ length of the zigzag scan is$.\; 1.\; 6$ sequences:1 5 3 9 7 3 9 5 4 7 3 6 6 4 1 3。

Please implement a $Z$ zigzag scanning procedure, a given$n\times n$ matrix, outputZZfor this matrix$Z$ result zigzag scan.

Input format
The first line of input contains an integer $n$ represents the size of the matrix.

Input the second line to the $n+1$ line each line contains$n$ positive integers, separated by spaces, represent a given matrix.

Output format
Output one line, including $n\times n$ integers, separated by spaces, indicating that the input matrix passes through$Z$ -shape results after scanning.

Data range
$1\le n\le 500,\; the$
matrix elements are not more than$1000\; is$ a positive integer.

Input sample:

4
1 5 3 9
3 7 5 6
9 4 6 4
7 3 1 3

Sample output:

1 5 3 9 7 3 9 5 4 7 3 6 6 4 1 3

Ideas:

As we all know, a two-dimensional matrix has a total of $2n-1$ positive diagonal and$2n-1$ sub-diagonal line.
And here we find that the scans are all along the sub-diagonal line. The picture below is to scan a$5\times 5$ Illustration of the matrix:

We found that the scan can be divided into two directions,右上and左下.

And because $n\times n$ matrix, all diagonal sub45degrees, i.e. the slope is1, we havey = x + b;i.e.x - y = b;wherebis the intercept, is fixed, i.e., the difference between the coordinates of the sub-diagonals is fixed.

Therefore, this relationship can be used to calculate coordinates.

Further, when we give the above figure 2n - 1diagonals from 0 ~ 2n - 2time number, you will find that the even-numbered diagonal down the right, and diagonal left is an odd-numbered laid down, as shown below:

Of course, when we scan, we should also ensure that the subscript does not cross the boundary.

Java code

import java.util.Scanner;

public class Main {
    
    
    public static void main(String[] args) {
    
    
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[][] arr = new int[n][n];
        for (int i = 0; i < n; i++) {
    
    
            for (int j = 0; j < n; j++) {
    
    
                arr[i][j] = sc.nextInt();
            }
        }

        //注意这里i表示的是当前遍历的是第几条副对角线，而非横坐标，共2n - 1条副对角线
        for (int i = 0; i <= 2 * n - 1; i++) {
    
     
            if ((i & 1) == 0) {
    
    //第偶数条对角线往右上走
                for(int j = i; j >= 0;j--){
    
    //j表示当前副对角线起点的横坐标，纵坐标由i - j算出即可
                    if (j >= 0 && j < n && i - j >= 0 && i - j < n) {
    
     //坐标越界检查
                        System.out.print(arr[j][i - j] + " ");
                    }
                }
            }else{
    
    //第奇数条对角线往左下走
                for (int j = 0; j <= i; j++) {
    
    //j表示当前副对角线起点的横坐标，纵坐标由i - j算出即可
                    if (j >= 0 && j < n && i - j >= 0 && i - j < n) {
    
     //坐标越界检查
                        System.out.print(arr[j][i - j] + " ");
                    }
                }
            }
        }
    }
}