4 lines in Java


  • 155

    Just checking the pairs, going inwards from the ends.

    public boolean isStrobogrammatic(String num) {
        for (int i=0, j=num.length()-1; i <= j; i++, j--)
            if (!"00 11 88 696".contains(num.charAt(i) + "" + num.charAt(j)))
                return false;
        return true;
    }

  • 0
    This post is deleted!

  • -2
    Y

    Well, this is smart solution.
    But how about the input string is "1=" or "8=" or "=1", etc, where "=" means space. It will return true but actually it is false. The test cases doesn't include above cases.
    I know this input might be invalid but the program is not robust enough..
    Just my 2 cents.


  • 6

    That's indeed invalid input. We're not asked to determine whether an arbitrary string represents a strobogrammatic number. We're asked to determine if a number is strobogrammatic. Which for technical reasons ends up as determining whether a string representing a number represents a strobogrammatic number.

    And if you think we should handle invalid input, then you should call your own three solutions "not robust enough" as well. I can make all of them crash.


  • 0
    Y

    Thanks for the comments. You're correct.


  • 0
    M

    I thought I know for loops lol. Neat solution


  • 0
    J

    This is really smart and clean, I like this!!


  • 0

    You can always think about the concise solution, smart!


  • 0
    L

    Wow you are really smart


  • 0
    A

    @StefanPochmann Hi, is the time complexity of this code O(N)?
    As the size of matcher string is always constant


  • 0

    Much smarter than using hashmap. Thanks for sharing!


  • 0
    F

    Amazing solution! Thanks for sharing!


  • 3

    The pairs are actually 00 11 69 96 88

    696 looks smart, but a little bit confusing to read


  • 0
    S

    Grrrrreat~ do not forget add validation before circle, which may cause NPE as num is null.


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