交际差集并集 java

//求两个字符串数组的并集，利用set的元素唯一性
public static String[] union(String[] arr1, String[] arr2) {
    Set<String> set = new HashSet<String>();
    for (String str : arr1) {
        set.add(str);
    }
    for (String str : arr2) {
        set.add(str);
    }
    String[] result = {};
    return set.toArray(result);
}

//求两个数组的交集
public static String[] intersect(String[] arr1, String[] arr2) {
    Map<String, Boolean> map = new HashMap<String, Boolean>();
    LinkedList<String> list = new LinkedList<String>();
    //通过hashmap key的特性,将重复的key设置为true,遍历key就可以了
    for (String str : arr1) {
        if (!map.containsKey(str)) {
            map.put(str, Boolean.FALSE);
        }
    }
    for (String str : arr2) {
        if (map.containsKey(str)) {
            map.put(str, Boolean.TRUE);
        }
    }

    for (Map.Entry<String, Boolean> e : map.entrySet()) {
        if (e.getValue().equals(Boolean.TRUE)) {
            list.add(e.getKey());
        }
    }

    String[] result = {};
    return list.toArray(result);
}

//求两个数组的差集
public static String[] minus(String[] arr1, String[] arr2) {
    LinkedList<String> list = new LinkedList<String>();
    LinkedList<String> history = new LinkedList<String>();
    String[] longerArr = arr1;
    String[] shorterArr = arr2;
    //找出较长的数组来减较短的数组
    if (arr1.length > arr2.length) {
        longerArr = arr2;
        shorterArr = arr1;
    }
    for (String str : longerArr) {
        if (!list.contains(str)) {
            list.add(str);
        }
    }
    for (String str : shorterArr) {
        if (list.contains(str)) {
            history.add(str);
            list.remove(str);
        } else {
            if (!history.contains(str)) {
                list.add(str);
            }
        }
    }

    String[] result = {};
    return list.toArray(result);
}