260. Single Number III 

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class Solution {
    public int[] singleNumber(int[] nums) {
        int xor = 0;
        for (int num : nums) {
            xor ^= num;
        }

        int mask = Integer.lowestOneBit(xor);

        int[] res = new int[2];
        for (int num : nums) {
            if ((mask & num) != 0) {
                res[0] ^= num;
            } else {
                res[1] ^= num;
            }
        }
        return res;
    }
}

References

260. Single Number III
剑指 Offer 56 - I. 数组中数字出现的次数