十亿手机号去重代码实践

思路：使用Java自带BitSet函数

public class Main {
    public static void main(String[] args) {
        BitSet bitSet = new BitSet();
        Map<Integer,BitSet> map=new HashMap<>();

        long[] phone={15522222222L,15533333333L,15544444444L,15522222222L};
        BitSet bitSet155 =map.computeIfAbsent(155,k->bitSet);
        // int num0= (int) (num/100000000);
        // System.out.println("前三位:"+num0);
        // int num1= (int) (num%100000000);
        // System.out.println(num1);

        for(int i=0;i<phone.length;i++){
            //int num0=(int) (phone[0]/100000000);
            int num1=(int) (phone[i]%100000000);
            bitSet155.set(num1);
        }
        for(int i=0;i<999999999;i++){
            if(bitSet155.get(i)){
                System.out.println("155"+i);
            }
        }
    }
}

截图：