如何將最小三角形邊的條件放到這個邊界上並獲得最小值的答案

Question

正如你看到這個程序應該給我多少行。 我想知道如何把一個條件給我最小的三角形邊 就好像我給數字5到（線）它會循環五次，每一行 我必須把3個分開的數字，但我不能找到一種方法來給我一個條件，讓他們中的哪一個是最小的一個。

import java.util.Scanner; 

public class Triangles { 
private int side[]; 

public static void main(String args[]) { 
    double line; 
    Scanner s = new Scanner(System.in); 
    System.out.println("Enter the number of Triangles:"); 
    line = s.nextInt(); 

    int side[] = new int[(int) line]; 

    System.out.println("Please, insert lengths of the sides of this triangles(3 real numbers per line) "); 

    for (int i = 0; i < side.length; i++) {// for reading array 
     side[0] = (int) s.nextDouble(); 
     side[1] = (int) s.nextDouble(); 
     side[2] = (int) s.nextDouble(); 
        double perimeter = side[0]+side[1]+side[2]; 
        System.out.println(perimeter); 
        System.out.println("Enter the next Triangles:"); 
    } 
    line--; 

} 
} 

Answer 1

根據您的要求，我修改了它。
我希望這將解決您的問題，請讓我知道如果您需要任何幫助

public static void main(String[] args) { 
     double line; 
     Scanner s = new Scanner(System.in); 
     System.out.println("Enter the number of Triangles:"); 
     line = s.nextInt(); 

     int side[] = new int[3];//because every triangle has 3 sides 
     double min = Double.MAX_VALUE;//assuming perimeter of a tringle will be grater than zero 
     while(line-->0){//we need to loop as many no Triangles 
      System.out.println("Please, insert lengths of the sides of this triangles(3 real numbers per line) "); 
     // for (int i = 0; i < side.length; i++) {// for reading array 
      side[0] = (int) s.nextDouble(); 
      side[1] = (int) s.nextDouble(); 
      side[2] = (int) s.nextDouble(); 
         double perimeter = side[0]+side[1]+side[2]; 
         if(perimeter<min){ 
          min = perimeter; 
         } 
         System.out.println(perimeter); 
         //System.out.println("Enter the next Triangles:"); 
     //} 
     } 
     System.out.println("Minimum Perimeter :"+min); 
     // line--; 

    }