java 代码

class Solution {
   public int findNumberAppearingOnce(int[] nums) {
        int[] bitSum = new int[32];  //int在二进制下是32位
        for (int num : nums) {  //外层循环遍历数组的每个数
            int bitMask = 1;  //数和bitMask & ,则对应位不为1就为0
            for (int j = 31; j >= 0; j--) {  //判断每个数的32个位置上的数是1还是2
                int bit = num & bitMask;
                if (bit != 0) bitSum[j] += 1;  //若为1则对应加上1
                bitMask = bitMask << 1;
            }
        }
        StringBuilder s1 = new StringBuilder();
        for (int i = 0; i < 32; i++) 
            s1.append(bitSum[i] % 3);
        String s2 = s1.toString();
        return Integer.parseInt(s2, 2);  //二进制字符串转化为十进制整数
    }
}