50.md

Pow(x,n)

Description

link

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Solution

See Code

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Code

O(logn) ~ O(log^2(n))

Iterative:

class Solution:
    def myPow(self, x, n):
        """
        :type x: float
        :type n: int
        :rtype: float
        """
        ans = 1.0
        m = abs(n)
        
        while m:
            if m & 1:
                ans *= x
            m >>= 1
            x *= x
            
        return ans if n > 0 else 1/ans


Recrusive:

class Solution:
    def myPow(self, x, n):
        """
        :type x: float
        :type n: int
        :rtype: float
        """
        if not n:
            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 >> 1)