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
转载于:https://www.cnblogs.com/vercont/p/10210380.html
