剑指offer——17数值的整数次方

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。

保证base和exponent不同时为0

一般解法：

　　直接相乘；

 1 class Solution01 {
 2 public:
 3     double Power(double base, int exponent) {
 4         if (exponent == 0 || equal(base, 0.0, 0.0))return 1;
 5         double res = 1.0;
 6         for (int i = 1; i <= abs(exponent); ++i)
 7             res *= base;
 8         return exponent < 0 ? (1.0 / res) : res;
 9     }
10 };

高效率解法：

　　使用递归：

 1 class Solution02 {
 2 public:
 3     double Power(double base, int exponent) {
 4         if (exponent == 0 || equal(base, 0.0, 0.0))return 1;
 5         double res = PowerMul(base, abs(exponent));
 6         return exponent < 0 ? (1.0 / res) : res;
 7     }
 8     double PowerMul(double base, unsigned int exponent)
 9     {
10         if (exponent == 0)return 1;
11         if (exponent == 1)return base;
12 
13         double res = PowerMul(base, exponent >> 1);
14         res *= res;
15         if (exponent & 1)//奇数
16             res *= base;
17         return res;
18     }
19 };