Pow(x, n) Leetcode Solution

In this post, we are going to solve the Pow(x, n) Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Pow(x, n) Leetcode Solution
Pow(x, n) Leetcode Solution

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
  • n is an integer.
  • -104 <= xn <= 104

Now, lets see the leetcode solution of Pow(x, n) Leetcode Solution.

Pow(x, n) Leetcode Solution 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)

Pow(x, n) Leetcode Solution in CPP

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) Leetcode Solution 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);
  }
}

Note: This problem Pow(x, n) is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

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