LeetCode 137: Single Number 2

public class SingleNumberII137 {
    /*
    problem: Given an array of integers, every element appears three
    times except for one, which appears exactly once. Find that single one.

    Solution: The usual bit manipulation code is bit hard to get and replicate.
    I like to think about the number in 32 bits and just count how many 1s are there
    in each bit, and sum %= 3 will clear it once it reaches 3. After running for all
     the numbers for each bit, if we have a 1, then that 1 belongs to the single number,
     we can simply move it back to its spot by doing ans |= sum << i;

   This has complexity of O(32n), which is essentially O(n) and very easy to think and
   implement. Plus, you get a general solution for any times of occurrence. Say all the
   numbers have 5 times, just do sum %= 5.
     */
    public int singleNumber(int[] nums) {
        int ans = 0;
        for(int i = 0; i < 32; i++) {
            int sum = 0;
            for(int j = 0; j < nums.length; j++) {
                if(((nums[j] >> i) & 1) == 1) {
                    sum++;
                    sum %= 3;
                }
            }
            if(sum != 0) {
                ans |= sum << i;
            }
        }
        return ans;
    }
}