劍指Offer 面試題11 數值的整數次方

實現double power（double base， int exponent），求base的exponent次方。不得使用庫函式，同時不需要考慮大數問題。

分析：

源**如下：

#includeusing std::cout;

using std::cin;
using std::endl;
class throwerror{};
double powerwithunsigned1(double base, int exponent)
double power1(double base, int exponent)
;		}
else
//指數大於0時，返回0
return 0;
}	else
else
}}void test11()
void test12()
void test13()
void test14()
void test15()
int main()

0^5 = 0

2^-5 = 0.03125
-2^-5 = -0.03125
-2^5 = -32
0^0 = 0
請按任意鍵繼續. . .

源**如下：

#includeusing std::cout;

using std::cin;
using std::endl;
class throwerror{};
double powerwithunsigned2(double base, int exponent)
double power2(double base, int exponent)
;		}
else
//指數大於0時，返回0
return 0;
}	else
else
}}void test21()
void test22()
void test23()
void test24()
void test25()
int main2()

注意：我們用右移替代了除以2，用位於運算替代了求餘運算子判斷是基數還是偶數，這種方法可以提高效率！

#include#include #includebool g_invalidinput = false;

bool equal(double num1, double num2);
double powerwithunsignedexponent(double base, unsigned int exponent);
double power(double base, int exponent)
unsigned int ab***ponent = (unsigned int)(exponent);
if (exponent < 0)
ab***ponent = (unsigned int)(-exponent);
double result = powerwithunsignedexponent(base, ab***ponent);
if (exponent < 0)
result = 1.0 / result;
return result;}/*
double powerwithunsignedexponent(double base, unsigned int exponent)
*/double powerwithunsignedexponent(double base, unsigned int exponent)
bool equal(double num1, double num2)
// ********************測試**********************
void test(double base, int exponent, double expectedresult, bool expectedflag)
int main()

test1 begins.

test passed.
test2 begins.
test passed.
test3 begins.
test passed.
test4 begins.
test passed.
test5 begins.
test passed.
test6 begins.
test passed.
test7 begins.
test passed.
請按任意鍵繼續. . .

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