C++ solution compare double and its floor

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    C++ solution compare double and its floor(or ground) number. float is ok in this problem.

    class Solution {
        bool isPowerOfThree(int n) {
            if(n <= 0) return false;
            double x = log10(n)/log10(3);
            return x == floor(x);

  • 0

    Could you explain why do you choose floor? Don't quite catch this point.

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    floor of 2.3 is 2.0 and floor of -2.3 is -3.0, Therefore, no matter the number x is positive or not, I just want to determine that it's a integer(0 after point). If x is an integer, so x == floor(x) should return true.

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    this method is not always reliable under finite precision. I tried change log10(n)/log10(3) in your code to log(n)/log(3) and now it can't pass test cases.

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    Tried that in Java, didn't quite work (FP precision). So I had to change it a little bit:

    1. Don't use floor(), use round() (if your language / library has that, otherwise do it yourself).
    2. Don't compare for equality. Instead, find the smallest tolerance that works and use it. By experimenting, I figured that 1E-11 works quite nice for all possible powers of three within 32-bit range.

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    I am not sure if there is precision problem if translate it to Java, so round() might be better, at least make more sense. Thanks.

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    Java or not, comparing FP using equality is usually a very bad idea. I know one case when a number was not equal to itself because at some point it was internally converted by the compiler between 64 and 80 bit representation and comparison used all 80 bits.

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