[LeetCode] 137. Single Number II

public class Solution {
    public int singleNumber(int[] nums) {
        int[] count=new int[32];
        int ans=0;
        for(int i=0;i<32;i++){
            for(int j=0;j<nums.length;j++){
                count[i]+=(nums[j]>>i)&1;

            }
            ans|=(count[i]%3)<<i;//当前一位不为3的倍数，说明single number在这里为1
        }
        return ans;
    }
}