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函数平方pow(x,n)的求法

2016-01-28 15:19 429 查看
考虑到n个x相乘式子的对称关系，可以对上述方法进行改进，从而得到一种时间复杂度为O(logn)的方法，递归关系可以表示为pow(x,n) = pow(x,n/2)*pow(x,n-n/2)

func pow(x float64, n int) float64 {
if n == 0 {
return 1.0
}
if n < 0 {
return 1.0 / pow(x, -n)
}

half := pow(x, n>>1) //example: n = 3 3/2 = 1

if n%2 == 0 {
return half * half
} else {
return half * half * x
}

}

除了上述方法，这里还提到了一种十分巧妙并且快速的方法，原文描述如下：

Consider the binary representation of n. For example, if it is "10001011", then x^n = x^(1+2+8+128) = x^1 * x^2 * x^8 * x^128. Thus, we don't want to loop n times to calculate x^n. To speed up, we loop through each bit, if the i-th bit is 1, then we add x^(1 << i) to the result. Since (1 << i) is a power of 2, x^(1<<(i+1)) = square(x^(1<<i)). The loop executes for a maximum of log(n) times.

func my_pow(x float64, n int) float64 {
if n == 0 {
return 1.0
}
if n < 0 {
return 1.0 / my_pow(x, -n)
}
var result float64 = 1.0
for ; n > 0; n = n >> 1 {
if n&1 != 0 {
result *= x
}
x = x * x
}
return result
}

完整代码 :

package main

import (
"fmt"
)

func pow(x float64, n int) float64 {
if n == 0 {
return 1.0
}
if n < 0 {
return 1.0 / pow(x, -n)
}
half := pow(x, n>>1) //example: n = 3 3/2 = 1
if n%2 == 0 {
return half * half
} else {
return half * half * x
}
}

func my_pow(x float64, n int) float64 {
if n == 0 {
return 1.0
}
if n < 0 {
return 1.0 / my_pow(x, -n)
}
var result float64 = 1.0
for ; n > 0; n = n >> 1 {
if n&1 != 0 {
result *= x
}
x = x * x
}
return result
}

func main() {
fmt.Printf("pow result %v \n", my_pow(6.0, 2))
}
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