数值的整数次方 牛客网 剑指Offer

• 题目描述
• 给定一个double类型的浮点数base和int类型的整数exponent。求base的exponent次方

class Solution:

    #run:23ms memory:5728k

    def Power(self, base, exponent):

        flag = 0

        if base == 0:

            return False

        if exponent == 0:

            return 1

        if exponent < 0:

            flag = 1

        ret = 1

        absExponent = abs(exponent)

        ret = self.pow(base,absExponent)

        if flag == 1:

            ret = 1/ret

        return ret

    def pow(self,X,N):

        if N == 0:

            return 1

        if N == 1:

            return X

        if N % 2 == 0:

            return self.pow(X*X,N/2)

        else:

            return self.pow(X*X,N/2) * X

    #run:21ms memory:5856k

    def Power2(self,base,exponent):

        flag = 0

        if base == 0:

            return False

        if exponent == 0:

            return 1

        if exponent < 0:

            flag = 1

        ret = 1

        absExponent = abs(exponent)

        for i in range(absExponent):

            ret *= base

        if flag == 1:

            ret = 1/ret

        return ret