Manhattan distance sum - print diamond

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Welcome to the blogger's magic tricks column , Detailed explanation of connotation algorithm foundation and code template

Title Description ：

• Enter an odd number n, Output a by * Composed of n Step solid diamond .

  Input format
  An odd number n.

  Output format
  Output a by * Composed of n Step solid diamond .

  Refer to the output sample for specific format .

  Data range
  1≤n≤99
  sample input ：
  5
  sample output ：
  

• Title source ：https://www.acwing.com/problem/content/729/

Topic analysis ：

Law 1 ： Manhattan distance and ：

• Manhattan distance ：x / y The maximum value of the direction projection distance

• Manhattan distance and ：x Direction projection distance + y Direction projection distance

• The diamond ：

  The nature of the circle itself is that the distance from each point to the center of the circle is equal , This distance is under the root sign ((Δx)²+(Δy)²)

  In the array , Because of the accuracy , Reduce the distance to Manhattan distance and
  The same Manhattan distance and the collection of points form a diamond , The diamond is already very close
  But because of its different mathematical expressions , The resulting figure is also a diamond

Algorithm template :

• Manhattan distance and Manhattan rectangle

Code implementation ：

Law 1 ：

• Enter only odd numbers ：

import java.util.Scanner;
import java.lang.Math;
public class Main{
    
    public static void main(String[] args){
    
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        for(int i=0; i<n; i++){
    
            for(int j=0; j<n; j++){
    
                if (Math.abs(i - n/2)+Math.abs(j - n/2) <= n/2){
    
                    System.out.print("*");
                }else System.out.print(" ");
            }
            System.out.println();
        }
    }
}

• Odd even mixed input ：

import java.util.Scanner;
import java.lang.Math;
public class Main{
    
    public static void main(String[] args){
    
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = n;
        if (n % 2 == 0) m++;
        for(int i=0; i<m; i++){
    
            for(int j=0; j<m; j++){
     
                if (Math.abs(i - m/2)+Math.abs(j - m/2) <= m/2){
    
                        System.out.print("*");
                    }else System.out.print(" ");  
                
            }
            System.out.println();
        }
    }
}

Be careful ：

• For odd row diamonds ：

  Input n It is the number of rows and columns in a diamond , It's diamond in the middle again * Number

  The number of lines of the square on the outside of the diamond / The number of columns is odd , Midpoint is (n/2 , n/2),mahattan = max(abs(i - n/2), abs(j - n/2))

• For even rows of diamonds ：

  Input n For the even , The number of diamond rows and columns is n+1, Middle row * The number is also n+1

  The number of lines of the square on the outside of the diamond / The number of columns is odd , Midpoint is ((n+1/2), (n+1)/2),mahattan = max(abs(i - (n+1)/2), abs(j - (n+1)/2))

• Can we force the use of even Manhattan distance formula to solve ？

  The answer is that this question cannot , After all, drawing a diamond can only have odd rows and odd sequences （2*n + 1）,
  The midpoint compulsorily solved by the even Manhattan distance formula is not the midpoint , Distance is even more wrong .

• demonstration ： Enter the following code 4 The diamond drawn is

import java.util.Scanner;
import java.lang.Math;
public class Main{
    
    public static void main(String[] args){
    
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = n;
        if (n % 2 == 0) m++;
        for(int i=0; i<m; i++){
    
            for(int j=0; j<m; j++){
    
                // Even numbers are wrong to use Manhattan distance formula , The midpoint position and distance are wrong 
                if(n % 2 == 1){
    
                    if (Math.abs(i - n/2)+Math.abs(j - n/2) <= n/2){
    
                        System.out.print("*");
                    }else System.out.print(" ");    
                }else{
    
                    if ((int)Math.abs(i - (n-1)/2.0)+(int)Math.abs(j - (n-1)/2.0) <= n/2){
    
                        System.out.print("*");
                    }else System.out.print(" ");    
                }
            }
            System.out.println();
        }
    }
}