数组中数字出现的次数（Java实现）

public class E56NumbersAppearOnce {
    //数组中数字出现的次数

    /*问题一：两个只出现一次的数字+出现两次的其它数字*/
    public static void findNumbers(int[] numbers, int length) {
        if (numbers == null || length <= 1)
            return;
        //保存所有数字的异或运算结果
        int orResult = 0;
        for (int i = 0; i < length; i++) {
            orResult ^= numbers[i];
        }
        //异或结果从右至左第一个1的位置
        int moveSteps = getPositionBitIs1(orResult);
        int number1 = 0;
        int number2 = 0;
        for (int i = 0; i < length; i++) {
            if (bitIs1(numbers[i], moveSteps))
                number1 ^= numbers[i];
            else
                number2 ^= numbers[i];
        }
        System.out.printf("只出现一次的两个数字分别为：%d，%d。", number1, number2);
    }

    private static int getPositionBitIs1(int orResult) {
        //返回右移次数
        int moveSteps = 0;
        while ((orResult & 1) == 0 && moveSteps < 32) {
            orResult = orResult >> 1;
            moveSteps++;
        }
        return moveSteps;
    }

    private static boolean bitIs1(int number, int moveSteps) {
        for (int i = 0; i < moveSteps; i++)
            number = number >> 1;
        return (number & 1) == 1;
    }

    /*问题二：只出现一次的一个数字+出现三次的其它数字*/
    public static int findNumber(int[] numbers, int length) {
        if (numbers == null || length <= 0)
            throw new IllegalArgumentException("Invalid Input!");
        int[] bits = new int[32];
        int bitMark = 1;
        for (int i = 31; i >= 0; i--) {
            for (int j = 0; j < length; j++){
                if ((numbers[j] & bitMark) != 0)
                bits[i] += 1;
            }

            bitMark = bitMark << 1;
        }
        int targetNumber = 0;
        for (int i = 0; i < 32; i++){
            targetNumber = targetNumber << 1;
            targetNumber += (bits[i] % 3);
        }
        return targetNumber;
    }

    //测试用例z
    public static void main(String[] args) {
        int[] numbers1 = {1, 1, 2, 2, 3, 4, 5, 5};
        E56NumbersAppearOnce.findNumbers(numbers1, 8);
        int[] numbers2 = {1, 1, 1, 2, 2, 2, 3, 4, 4, 4};
        System.out.println("\n" + E56NumbersAppearOnce.findNumber(numbers2, 10));
    }
}