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用List接口创设一个20个100以内数字的

发布时间: 2012-09-07 10:38:15 作者: rapoo

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]

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

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