Niuke brushing notes-introductory training for programming beginners (simple article 3)

BC68-X pattern

import java.util.Collections;
import java.util.Scanner;
public class Main {
    
    
    public static void main(String[] args) {
    
    
        Scanner sc = new Scanner(System.in);
        while (sc.hasNext()) {
    
    
            int n = sc.nextInt();
            String str = String.join("",Collections.nCopies(n," "));
            for (int i = 0; i < n; i++) {
    
    
                char[] temp = str.toCharArray();
                temp[i] = '*';
                temp[n-i-1] = '*';
                System.out.println(new String(temp));
            }
        }
    }
}

I have done several questions about outputting patterns, and the code gets longer and longer. Refer to other people's code to learn new knowledge.

  1. List l = new ArrayList();// Using the ArrayList class to instantiate the List collectionInsert picture description here

  2. It is often necessary to concatenate the items of an array or List into a string through a separator. The general implementation logic is to connect through member + separator, and then cut off the last separator in the result. The String.Join method implements this function.

    List names=new ArrayList<String>();
    names.add("1");
    names.add("2");
    names.add("3");
    System.out.println(String.join("-", names));
    String[] arrStr=new String[]{
          
          "a","b","c"};
    System.out.println(String.join("-", arrStr));
    输出:
    1-2-3
    a-b-c
    
  3. The ncopies(int, T) method is used to return an immutable list of n copies of the specified object. Declaration: public static List nCopies(int n, T o)
    Parameters: n-- the number of elements in the returned list; o-- the repeated elements in the returned list. Return value: The specified object of n copies of the immutable list returned by the method call. Exception: IllegalArgumentException-- If n <0 this exception is thrown.

BC74-HTTP status code

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class Main {
    
    
    public static void main(String[] args) {
    
    
        Scanner sc = new Scanner(System.in);
        Map<Integer,String> map = new HashMap<>(16);
        map.put(200,"OK");
        map.put(202,"Accepted");
        map.put(400,"Bad Request");
        map.put(403,"Forbidden");
        map.put(404,"Not Found");
        map.put(500,"Internal Server Error");
        map.put(502,"Bad Gateway");
        while (sc.hasNext()) {
    
    
            int state = sc.nextInt();
            System.out.println(map.get(state));
        }
    }
}

After using switch to watch other people's method, I found this approach, and learn about map.

In the collection of Map, elements exist in pairs (understood as husband and wife). Each element is composed of two parts, a key and a value, and the corresponding value can be found through the key. The collection in the Map cannot contain duplicate keys, and the values ​​can be duplicated; each key can only correspond to one value.
HashMap<K,V>: The structure of the hash table used to store data. The access sequence of elements cannot be guaranteed to be consistent. To ensure the uniqueness and non-repetition of the key, the hashCode() method and equals() method of the key need to be rewritten.
LinkedHashMap<K,V>: There is a subclass LinkedHashMap under HashMap, which uses the hash table structure + linked list structure to store data. The linked list structure can ensure that the access sequence of the elements is consistent; the unique and non-repetitive keys that can be ensured by the hash table structure, the hashCode() method and equals() method of the key need to be rewritten.
Note: The collection in the Map interface has two generic variables <K,V>. When using, the two generic variables must be assigned data types. The data types of the two generic variables <K,V> can be the same or different.

BC80-Login Verification

import java.util.Scanner;
public class Main {
    
    
    public static void main(String[] args) {
    
    
    Scanner sc=new Scanner(System.in);
    String[] strs=sc.nextLine().split(" ");
    if(strs[0].equals("admin")&&strs[1].equals("admin")) {
    
    
         System.out.println("Login Success!");
        }else {
    
    
         System.out.println("Login Fail!");
        }
    }
}

Method for judging the same string: String.equals()

BC85-a number containing the number 9

public class Main {
    
    
    public static void main(String[] args) {
    
    
     int sum=0;
     for(Integer i=1;i<=2019;i++) {
    
    
        if(i.toString().contains("9")) {
    
    
            sum++;
        }
     }
     System.out.println(sum);
    }
}
  1. toString() has two forms: toString() returns a String object representing an Integer value; toString(int i) returns a String object representing a specified int.
  2. contains() determines whether there is a substring in the string. It returns true if it has, and false if it doesn't.
  3. Integer is a wrapper class of int, int is a basic data type of java;
    Integer variables must be instantiated before they can be used, and int variables are not needed;
    Integer is actually a reference to an object, when new is an Integer, it is actually generated A pointer points to this object; while int stores the data value directly;
    the default value of Integer is null, and the default value of int is 0.

other

import java.util.Arrays;
public class Main {
    
    
     public static void main(String[] args) {
    
    
         int[] a = {
    
    9, 8, 7, 2, 3, 4, 1, 0, 6, 5};
         Arrays.sort(a);
         for(int i = 0; i < a.length; i ++) {
    
    
             System.out.print(a[i] + " ");
         }
     }
}
输出:
0 1 2 3 4 5 6 7 8 9

1. Arrays.sort(int[] a)
This form is to sort all the elements of an array, and in the order from small to large.
2. Arrays.sort(int[] a, int fromIndex, int toIndex)
This form is to sort the array part, that is, to sort the elements of the subscript of array a from fromIndex to toIndex-1. Note: the subscript is toIndex The elements do not participate in sorting.

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Origin blog.csdn.net/qq_41536271/article/details/112337146