使用正则表达式检查 int/string 是否包含特定数字

Question

I have a string/integer X: 912035374356789

i want to check if X contains: 1, 2, 3, 4, 5, 6, 7, 8, 9, 0

occurrence order is not matter

How do check it using regular expression. if there is any algorithm please mention, because i want to do it in minimum time complexity

Example
ex1: 61243456092 //false 7 & 8 not present
ex2: 864123456789 //false 0 is not present
ex3: 987601234567 //true all present

Answer 1

You can use the following regex blocks that ensure 1 is there at least once.

(?=.*1)

Now, in your case, you can combine all of them (positive look ahead)

(?=.*0)(?=.*1)(?=.*2)(?=.*3)(?=.*4)(?=.*5)(?=.*6)(?=.*7)(?=.*8)(?=.*9)

Demo: demo

Answer 2

If the string contains only digits then you can count the number of unique digits in it

long count = string.chars().distinct().count();

and check if count is 10

Example

String ex1 = "61243456092";
String ex2 = "864123456789";
String ex3 = "987601234567";

System.out.println(ex1.chars().distinct().count());
System.out.println(ex2.chars().distinct().count());
System.out.println(ex3.chars().distinct().count());

yields

8
9
10

Answer 3

I would just use string.contains like this:

public static boolean checkString(String str) {
    for (int i = 0; i < 10; i++) {
        if (!str.contains(Integer.toString(i))) {
            return false;
        }
    }

    return true;
}

I realize it's not a regular expression like you wanted, but I think it's the simplest answer.