137. 只出现一次的数字 II

class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        bits = [0] * 32
        for num in nums:
            for i in range(32):
                bits[i] += (num & 1)
                num >>= 1
        res = 0
        for i in range(32):
            if bits[i] % 3 != 0:
                res |= (1 << i)
        # 如果为负数，先将 0-32 位取反（即 res ^ 0xffffffff ），再将所有位取反（即 ~ ）
        return res if bits[31] % 3 == 0 else ~(res ^ 0xffffffff)

class Solution {
    public int singleNumber(int[] nums) {
        int[] bits = new int[32];
        for (int num : nums) {
            for (int i = 0; i < 32; ++i) {
                bits[i] += (num & 1);
                num >>= 1;
            }
        }

        int res = 0;
        for (int i = 0; i < 32; ++i) {
            if (bits[i] % 3 == 1) {
                res |= (1 << i);
            }
        }
        return res;
    }
}