50. Pow(x, n)

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

My Answer 1: Accepted (Runtime: 32 ms - 60.80% / Memory Usage: 14.3 MB - 19.59%)

class Solution:
    def myPow(self, x: float, n: int) -> float:
        if n > 0:
            return x**n
        else:
            n *= -1
            return (1/x)**n

양수면 x 의 n 승 반환, 음수면 1/x 의 -n 승 반환

반복문으로 하나하나 곱해주는게 가장 느렸다..
이거도 올리면 극딜 당할 거 같고.. 뭔가 규칙이 있을 거 같은디

Solution 1: Runtime: 32 ms - 60.80% / Memory Usage: 14.3 MB - 53.88%

class Solution:
    def myPow(self, x: float, n: int) -> float:
        # x**n == (1/x)**(-n) 
        # by using the property above we can transform the negetive power problem to positive power problem
        # so that we solve the positive power situation, we also solved the negtive power situation.
        if n < 0:
            x = 1/x
            n = -n
        # We solve the positive power here:
        power = 1
        current_product = x
        while n > 0:
            # if n is odd numberm, we need to time x one more time
            if n%2 : 
                power = power * current_product
            current_product = current_product * current_product
            n = n//2
        return power

• x^n = (1/x)^(-n)
• x^(2n) = (x^n) *(x^n)

n 을 2 로 나눠가면서 곱함

Recursive Solution

Solution 2: Runtime: 32 ms - 60.80% / Memory Usage: 14.4 MB - 19.59%

class Solution:
    def myPow(self, x: float, n: int) -> float:

        def function(base = x, exponent = abs(n)):
            if exponent == 0:
                return 1
            elif exponent % 2 == 0:
                return function(base * base, exponent // 2)
            else:
                return base * function(base * base, (exponent - 1) // 2)

        f = function()
        
        return float(f) if n >= 0 else 1/f

재귀다.

홀짝 여부에 따라 변환해서 푸는군요..