LeetCode Solutions
50. Pow(x, n)
Time: $O(\log n)$ Space: $O(1)$
class Solution {
public:
double myPow(double x, long n) {
if (n == 0)
return 1;
if (n < 0)
return 1 / myPow(x, -n);
if (n & 1)
return x * myPow(x, n - 1);
return myPow(x * x, n / 2);
}
};
class Solution {
public double myPow(double x, long n) {
if (n == 0)
return 1;
if (n < 0)
return 1 / myPow(x, -n);
if (n % 2 == 1)
return x * myPow(x, n - 1);
return myPow(x * x, n / 2);
}
}
class Solution:
def myPow(self, x: float, n: int) -> float:
if n == 0:
return 1
if n < 0:
return 1 / self.myPow(x, -n)
if n & 1:
return x * self.myPow(x, n - 1)
return self.myPow(x * x, n // 2)