leetcode — single-number-ii

/**
 * Source : https://oj.leetcode.com/problems/single-number-ii/
 *
 * Given an array of integers, every element appears three times except for one. Find that single one.
 *
 * Note:
 * Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
 *
 */
public class SingleNumber2 {

    /**
     * 数组中每个元素出现3次，只有一个元素出现一次，找出出现一次的元素
     *
     * 沿用SingleNumber的方法，比如：{1,1,1,2,2,2,3}
     * 01
     * 01
     * 01
     * 10
     * 10
     * 10
     * 11
     * ______
     * 44 对每一位求和
     * 11 每一位对3求余
     *
     * @param arr
     * @return
     */
    public int singleNumber (int[] arr) {
        int res = 0;
        for (int i = 0; i < 32; i++) {
            int sum = 0;
            int mask = 1<<i;
            for (int j = 0; j < arr.length; j++) {
                if ((arr[j] & mask) != 0) {
                    sum ++;
                }
            }
            res |= (sum % 3) << i;
        }
        return res;
    }

    public static void main(String[] args) {
        SingleNumber2 singleNumber2 = new SingleNumber2();
        System.out.println(singleNumber2.singleNumber(new int[]{1,1,1,2,2,2,3}) + " == 3");
        System.out.println(singleNumber2.singleNumber(new int[]{14,14,14,9}) + " == 9");
    }

}