剑指Offer 16 数值的整数次方

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。

 1 # -*- coding:utf-8 -*-
 2 class Solution:
 3     def Power(self, base, exponent):
 4         if exponent == 0:
 5             return 1
 6         if exponent == 1:
 7             return base
 8         op = False
 9         if exponent < 0:
10             op = True
11             exponent = -exponent
12         result = 1
13         if exponent % 2 == 0:
14             result = self.Power(base * base,exponent // 2)
15         else:
16             result = self.Power(base * base,exponent // 2) * base
17         if op:
18             result = 1 / result
19         return result
20         # write code here

leetcode地址，Java版代码：

 1 class Solution {
 2     public double myPow(double x, int n) {
 3         if (n == 0) {
 4             return 1;
 5         }
 6         if (n == 1) {
 7             return x;
 8         }
 9         if(n == -1){
10             return 1 / x;
11         }
12 
13         double half = myPow(x, n / 2);
14         double mod = myPow(x, n % 2);
15         return half * half * mod;
16     }
17 }

posted on 2019-06-12 22:11 Sempron2800+ 阅读(114) 评论(0) 编辑收藏举报