對List去重並保證新增順序主要有三種方式：

方式一，利用HashSet不能新增重複資料的特性 由於HashSet不能保證新增順序，所以只能作為判斷條件：

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不能新增重複資料並能保證新增順序的特性 ：

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方法迴圈遍歷：

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);
}
 

準備測試程式：

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));

}
 

結果為：

time1:329
time2:292
time3:17315