用List接口创建一个20个100以内数字的有序集合

import java.util.ArrayList;
import java.util.List;
import java.util.Random;

 

public class Test2 {

 public static void main(String[] args) {

  Random ran = new Random();

  List list = new ArrayList();

  while (true) {

   int n = ran.nextInt(100);

   if (!list.contains(n)) {
    list.add(n);
   }
   if (list.size() == 20) {
    break;
   }
  }
  

// 排序

  System.out.println(list);

  for (int i = 0; i < list.size() - 1; i++) {

   for (int j = i + 1; j < list.size(); j++) {

    int n1 = Integer.parseInt(list.get(i).toString());
    int n2 = Integer.parseInt(String.valueOf(list.get(j)));

    if (n1 > n2) {

     list.set(i, n2);
     list.set(j, n1);
    }

   }

  }
  
  System.out.println(list);
 }
}

 

打印结果：

[70, 89, 80, 12, 2, 40, 57, 9, 4, 32, 42, 47, 36, 1, 15, 21, 49, 29, 38, 45]
[1, 2, 4, 9, 12, 15, 21, 29, 32, 36, 38, 40, 42, 45, 47, 49, 57, 70, 80, 89]

评论

1 楼 jyjava 2011-12-17  

大哥你冒泡啊，还不如转为数组，然后排序结束，再转回List