50. Pow(x,n)

Implement pow(x, n), which calculates x raised to the power n (i.e., \(x^n\)).

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\(^{-2}\) = 1/2\(^{2}\) = 1/4 = 0.25

Solution:

class Solution(object):
    def myPow(self, x, n):
        """
        :type x: float
        :type n: int
        :rtype: float
        """
        if n == 0:
            return 1.0
        if n < 0:
            x = 1 / x
            n = -n
        result = 1
        while n > 0:
            if n % 2 == 1:
                result *= x
            x *= x
            n //= 2
        return result

Time Complexity:

The time complexity of this function is O(log n), where n is the exponent n. This is because the algorithm divides the exponent by 2 at each iteration of the while loop until it reaches 0, which takes log(n) iterations. Each iteration takes constant time.

Space Complexity:

The space complexity of this function is O(1) because the function only uses a constant amount of extra space to store variables regardless of the size of the input.