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Link for the Problem – Pow(x, n)– LeetCode Problem
Pow(x, n)– LeetCode Problem
Problem:
Implement pow(x, n), which calculates x
raised to the power n
(i.e., xn
).
Example 1:
Input: x = 2.00000, n = 10 Output: 1024.00000
Example 2:
Input: x = 2.10000, n = 3 Output: 9.26100
Example 3:
Input: x = 2.00000, n = -2 Output: 0.25000 Explanation: 2-2 = 1/22 = 1/4 = 0.25
Constraints:
-100.0 < x < 100.0
-231 <= n <= 231-1
-104 <= xn <= 104
Pow(x, n)– LeetCode Solutions
Pow(x, n) in C++:
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); } };
Pow(x, n) in Java:
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); } }
Pow(x, n) in Python:
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 % 2: return x * self.myPow(x, n - 1) return self.myPow(x * x, n / 2)
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