leetcode-914-easy

X of a Kind in a Deck of Cards
思路一: 统计数字个数，然后用 [2, length] 的约数尝试所有数字个数，判断能否整除

public boolean hasGroupsSizeX(int[] deck) {
    Map<Integer, Integer> map = new HashMap<>();

    for (int val : deck) {
        map.compute(val, (k, v) -> v == null ? 1 : v + 1);
    }

    for (int X = 2; X <= deck.length; ++X) {
        if (deck.length % X == 0) {
            boolean flag = true;
            for (Integer value : map.values()) {
                if (value % X != 0) {
                    flag = false;
                    break;
                }
            }
            if (flag) {
                return true;
            }
        }
    }

    return true;
}