更简洁的方法来检查ArrayList中的Integers?

时间:2017-05-15 15:59:50

标签: java android list arraylist

是否有更短的替代方法来顺序检查ArrayList中的多个&&条件?

if (mList.contains(0) && mList.contains(1) && mList.contains(2)) {
    // TODO: Do I really have to do it like this?
};

或者我可以这样做吗?

    Integer[] condition1 = {0, 1, 2};
        for (Integer item : condition1) {
            if (mXPositions.containsAll(item)) {
                Toast.makeText(mContext, R.string.player_1_wins, Toast.LENGTH_SHORT).show();
            } else if (mOPositions.contains(item)) {
                Toast.makeText(mContext, R.string.player_2_wins, Toast.LENGTH_SHORT).show();
            }
     }

1 个答案:

答案 0 :(得分:3)

你可以通过Arrays.asList()缩短它,这会接受一个varagrs参数:

if (mList.containsAll(Arrays.asList(0, 1, 2)))
    // ...
}