原题链接: https://oj.leetcode.com/problems/single-number-ii/

public int singleNumberII(int[] A) {
    if (A == null || A.length == 0) return 0;

    // the solving key is to regard the number as binary number, not decimal number,
    // so when a num exists 'n' times , the bits of it will multiply 'n' time too.

    int len = A.length;
    int[] digits = new int[32];

    for (int i = 0; i < 32; i++) {
        for (int j = 0; j < len; j++) {
            digits[i] += A[j] >> i & 1;    // get the i-th bit of every number of A
        }
    }

    int result = 0;

    for (int i = 0; i < 32; i++) {
        result += digits[i] % 3 << i;
    }

    return result;
}