math - 如何找到具有相同二进制子串的下一个整数？

Answer 1

Not sure exactly what you are trying to accomplish here, but bit shift left is the same as saying multiply by two. So if you want to get the "next" integer with the same "substring" you just multiply by two.

3: 11 
*2=
6: 110

18: 10010
*2=
36: 100100

Answer 2

Making a bit of jump on your methodology :P but you could convert the number to a string and use some sort of find method. It doesn't do it via math? but it gets the job done

public static void main(String[] args){

    Integer myNum = 1010;
    String myNumString = Integer.toString(myNum);
    System.out.println(myNumString.matches(".*10.*"));
}