用C语言求一个整数的幂的最有效方法是什么?
// 2^3
pow(2,3) == 8
// 5^5
pow(5,5) == 3125
用C语言求一个整数的幂的最有效方法是什么?
// 2^3
pow(2,3) == 8
// 5^5
pow(5,5) == 3125
当前回答
我的情况有点不同,我试图用一种力量创造一个面具,但我想无论如何我都要分享我找到的解决方案。
显然,它只适用于2的幂。
Mask1 = 1 << (Exponent - 1);
Mask2 = Mask1 - 1;
return Mask1 + Mask2;
其他回答
int pow(int const x, unsigned const e) noexcept
{
return !e ? 1 : 1 == e ? x : (e % 2 ? x : 1) * pow(x * x, e / 2);
//return !e ? 1 : 1 == e ? x : (((x ^ 1) & -(e % 2)) ^ 1) * pow(x * x, e / 2);
}
是的,它是递归的,但是一个好的优化编译器会优化递归。
如果你想得到一个整数的2的幂,最好使用shift选项:
Pow(2,5)可以替换为1<<5
这样效率更高。
更一般的解决方案考虑负指数
private static int pow(int base, int exponent) {
int result = 1;
if (exponent == 0)
return result; // base case;
if (exponent < 0)
return 1 / pow(base, -exponent);
int temp = pow(base, exponent / 2);
if (exponent % 2 == 0)
return temp * temp;
else
return (base * temp * temp);
}
O(log N)的解决方案在Swift…
// Time complexity is O(log N)
func power(_ base: Int, _ exp: Int) -> Int {
// 1. If the exponent is 1 then return the number (e.g a^1 == a)
//Time complexity O(1)
if exp == 1 {
return base
}
// 2. Calculate the value of the number raised to half of the exponent. This will be used to calculate the final answer by squaring the result (e.g a^2n == (a^n)^2 == a^n * a^n). The idea is that we can do half the amount of work by obtaining a^n and multiplying the result by itself to get a^2n
//Time complexity O(log N)
let tempVal = power(base, exp/2)
// 3. If the exponent was odd then decompose the result in such a way that it allows you to divide the exponent in two (e.g. a^(2n+1) == a^1 * a^2n == a^1 * a^n * a^n). If the eponent is even then the result must be the base raised to half the exponent squared (e.g. a^2n == a^n * a^n = (a^n)^2).
//Time complexity O(1)
return (exp % 2 == 1 ? base : 1) * tempVal * tempVal
}
下面是Java中的方法
private int ipow(int base, int exp)
{
int result = 1;
while (exp != 0)
{
if ((exp & 1) == 1)
result *= base;
exp >>= 1;
base *= base;
}
return result;
}