如何把数字以升序排列在Java中的ArrayList中？

Question

我已经做了一些代码，它从标准输入中获取数字，我想要将这些值按升序排序。我该怎么做？

import java.util.*; 

public class Main { 

    static Scanner sc = new Scanner(System.in); 
    public static void main(String[] args) { 
     boolean Done = false; 
     ArrayList<Integer> numbers = new ArrayList<Integer>(20); 


     while(Done == false){ 
      System.out.print("Enter number: "); 
      int x = sc.nextInt(); 
      numbers.add(x); 

      System.out.print("Keep going? (Y/N)"); 
      String keepGoing = sc.next(); 

      if(keepGoing.equalsIgnoreCase("Y")) 
       ; 
      else if(keepGoing.equalsIgnoreCase("N")){ 
       Done = true; 

           //this is just temporary to print out the numbers. 
           for(int n : numbers){ 
        System.out.print(n + " "); 
       } 
      } 
      else 
       System.out.println("Command not recognised..."); 
     } 



    } 
} 

很抱歉，如果我张贴已经被回答的问题，但我找不到这似乎与我的代码:(

Answer 1

Collections.sort(numbers);应该诀窍

Answer 2

只是sort工作的任何答案ArrayList：

Collections.sort(numbers); 

Answer 3

import java.util.*; 

public class Main { 

    static Scanner sc = new Scanner(System.in); 

    public static void main(String[] args) { 
     boolean Done = false; 
     ArrayList<Integer> numbers = new ArrayList<Integer>(20); 
     while (Done == false) { 
      System.out.print("Enter number: "); 
      int x = sc.nextInt(); 
      numbers.add(x); 

      while (true) { 
       System.out.print("Keep going? (Y/N) :"); 
       String keepGoing = sc.next(); 
       if (keepGoing.equalsIgnoreCase("Y")) 
        break; 
       else if (keepGoing.equalsIgnoreCase("N")) { 
        Done = true; 
        break; 
       } 
       System.out.println("Command not recognised..."); 
      } 
     } 

     System.out.println("before sort : " + numbers); 
     Collections.sort(numbers); 
     System.out.println("after sort : " + numbers); 
    } 
}