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贵有恒何必三更眠五更起 最无益只怕一日曝十日寒

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对List去重并保证添加顺序主要有三种方式:

方式一,利用HashSet不能添加重复数据的特性 由于HashSet不能保证添加顺序,所以只能作为判断条件:

private static void removeDuplicate(List<String> list) {
    HashSet<String> set = new HashSet<String>(list.size());
    List<String> result = new ArrayList<String>(list.size());
    for (String str : list) {
        if (set.add(str)) {
            result.add(str);
        }
    }
    list.clear();
    list.addAll(result);
}

方式二,利用LinkedHashSet不能添加重复数据并能保证添加顺序的特性 :

private static void removeDuplicate(List<String> list) {
    LinkedHashSet<String> set = new LinkedHashSet<String>(list.size());
    set.addAll(list);
    list.clear();
    list.addAll(set);
}

 

方式三,利用List的contains方法循环遍历:

private static void removeDuplicate(List<String> list) {
    List<String> result = new ArrayList<String>(list.size());
    for (String str : list) {
        if (!result.contains(str)) {
            result.add(str);
        }
    }
    list.clear();
    list.addAll(result);
}

准备测试程序:

private static void main(String[] args) {
    final List<String> list = new ArrayList<String>();
    for (int i = 0; i < 1000; i++) {
        list.add("haha-" + i);
    }

    long time = System.currentTimeMillis();
    for (int i = 0; i < 10000; i++) {
        removeDuplicate(list);
    }
    long time1 = System.currentTimeMillis();
    System.out.println("time1:"+(time1-time));

    for (int i = 0; i < 10000; i++) {
        removeDuplicate2(list);
    }
    long time2 = System.currentTimeMillis();
    System.out.println("time2:"+(time2-time1));

    for (int i = 0; i < 10000; i++) {
        removeDuplicate3(list);
    }
    long time3 = System.currentTimeMillis();
    System.out.println("time3:"+(time3-time2));

}
posted on 2018-10-24 19:23  白露~  阅读(2246)  评论(0编辑  收藏  举报