260. Single Number III

class Solution {
    public int[] singleNumber(int[] nums) {
        // 相同 bit 异或为 0, 不同 bit 异或为 1
        
        int mergedMask = 0;
        for (int num : nums) {
            mergedMask ^= num;
        }

        int diffMask = 1;
        while ((mergedMask & diffMask) == 0) {
            diffMask <<= 1;
        }

        int x = 0, y = 0;
        for (int num : nums) {
            if ((num & diffMask) != 0) {
                x ^= num;
            } else {
                y ^= num;
            }
        }

        return new int[]{x, y};
    }
}

Reference

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