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50. Powx N

Math Recursion

Problem - Powx N

Medium

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.
  • Either x is not zero or n > 0.
  • -104 <= xn <= 104

Solutions

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class Solution:
    def myPow(self, x: float, n: int) -> float:
        def powfunc(val: float, power: int) -> float:
            total = 1
            while power:
                if power & 1:
                    total *= val
                val *= val
                power >>= 1
            return total

        if n >= 0:
            return powfunc(x, n)
        else:
            return 1/ powfunc(x, -n)

Submission Stats:

  • Runtime: 0 ms (100.00%)
  • Memory: 17.9 MB (49.94%)