Java 如何在40亿个整数中寻找一个数字

前几天刷到一个视频漫画面试系列-如何在40亿个整数中寻找一个数字。作者只是给出了思路用bitmap判断，没有具体的代码实现。我自己思考了一下，给出解法：
教学视频

import java.util.HashSet;
import java.util.Set;

public class Billion42 {
    private Set set32_16, set24_16, set16_0;

    Billion42() {
        set32_16 = new HashSet<>();
        set24_16 = new HashSet<>();
        set16_0 = new HashSet<>();
    }

    public void add(int num) {
        short val32_16, val24_16, val16_0;
        val32_16 = (short) (num & 0xFFFF0000);
        val24_16 = (short) (num & 0x00FFFF00);
        val16_0 = (short) (num & 0x0000FFFF);

        if (!(set32_16.contains(val32_16) &&
                set24_16.contains(val24_16) &&
                set16_0.contains(val16_0))) {
            set32_16.add(val32_16);
            set24_16.add(val24_16);
            set16_0.add(val16_0);
        }
    }

    public boolean isExists(int num) {
        short val32_16, val24_16, val16_0;
        val32_16 = (short) (num & 0xFFFF0000);
        val24_16 = (short) (num & 0x00FFFF00);
        val16_0 = (short) (num & 0x0000FFFF);
        return (set32_16.contains(val32_16) &&
                set24_16.contains(val24_16) &&
                set16_0.contains(val16_0));
    }

}

public class main {

    public static void main(String[] args) {
        int rand = 0;
        Billion42 bill = new Billion42();

        for (long i = 0; i < 4294967295L; ++i) {
            rand = (int) (Math.random() * (1073741823 - 0) + 0);
            bill.add(rand);
        }
        int num = (int) (Math.random() * (1073741823 - 0) + 0);
        //bill.add(num);
        System.out.println(num);
        //拆分32位整数后拆分后重新生成
        System.out.println(
                (num & 0xFFFF0000) | (num & 0x00FFFF00) | (num & 0x0000FFFF)
        );
        System.out.println(bill.isExists(num));
    }
}

将整数按16bit进行拆分，32位整数拆分为3个部份（64位整数拆分成7个部份），将这3个部份存储至3个set中。判断一个数字是否存在时判断该数字的所有bit位是否相同，从而来判断数字是否存在。

//高位16bit
(num & 0xFFFF0000)
//中间位16bit
(num & 0x00FFFF00)
//低位16bit
(num & 0x0000FFFF)

(num & 0xFFFF000000000000)
(num & 0x00FFFF0000000000)
(num & 0x0000FFFF00000000)
(num & 0x000000FFFF000000)
(num & 0x00000000FFFF0000)
(num & 0x0000000000FFFF00)
(num & 0x000000000000FFFF)

16bit的最大值为65536，因此每个set里元素最多只会有65535个元素；short类型大小为2字节

所以一个set占用的内存大小为

65535  * 2 = 131070byte

然后再用set数量乘131070得到占用内容的大小

 * 拆分的数量 * 2字节short * 65535
#32位整数
65535 * 2 * 3 = 393210byte = 383kb
#64位整数
65535 * 2 * 7 = 917490byte = 895kb