Leetcode 50: Pow(x, n)

leetcode_50.cpp

#include <iostream>

using namespace std;

double myPow(double x, int n) {
    if (n==0)
        return (x<0) ? -1 : 1;
    long int e = n;  //one of the test cases uses INT_MIN, which exceeds range when doing * -1, so long int increases range
    if (e<0) {
        x = 1/x;
        e = e * -1;
    }

    double res = 1;
    while (e > 0) {
        if (e & 1)
            res = res * x;  //if exponent is odd then multiply res by our current factor of x

        //due to above, n is now effectively even
        e = e >> 1;   //bitshift n, i.e. divide exponent by 2
        x = x * x;    //square x, i.e. x -> x^2
    }
    return res;
}

int main() {
    cout << myPow(2.1, 3 ) << "\n";  //myPow2(0.00001, 2147483647) << "\n";  //myPow2(3,5) << "\n"; 

    return 0;
}