List去重的三种方式及效率对比

小咪咪 2022-05-17 00:05 168阅读 0赞

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));
    
    }

结果为：

time1:329
    time2:292
    time3:17315