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Power.java
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package Chap3;
/**
* 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。
* 不得使用库函数直接实现，无需考虑大数问题。
*/
public class Power {
/**
* 容易想到的蹩脚办法
*
* @param base 基数
* @param exponent 次幂
* @return base^exponent
*/
public double power_2(double base, int exponent) {
if (base == 0) {
return 0;
}
double result = 1.0;
int positiveExponent = Math.abs(exponent);
for (int i = 0; i < positiveExponent; i++) {
result *= base;
}
return exponent < 0 ? 1 /result : result;
}
/**
* 非递归。推荐的做法，复杂度O(lg n)
*/
public double power(double base, int exponent) {
if (base == 0) {
return 0;
}
double result = 1.0;
int positiveExponent = Math.abs(exponent);
while (positiveExponent != 0) {
// positiveExponent & 1这句是判断奇数的
if ((positiveExponent & 1) == 1) {
result *= base;
}
base *= base;
// 右移1位等于除以2
positiveExponent = positiveExponent >> 1;
}
return exponent < 0 ? 1 / result : result;
}
/**
* 和上面是同一个思路，递归版本。
*/
private double powerUnsigned(double base, int exponent) {
if (exponent == 0) {
return 1;
}
if (exponent == 1) {
return base;
}
double result = powerUnsigned(base, exponent >> 1);
result *= result;
if ((exponent & 1) == 1) {
result *= base;
}
return result;
}
public double power_1(double base, int exponent) {
if (base == 0) {
return 0;
}
int positiveExponent = Math.abs(exponent);
double result = powerUnsigned(base, positiveExponent);
return exponent < 0 ? 1 / result : result;
}
public static void main(String[] args) {
Power a = new Power();
System.out.println(a.power_2(2, -2));
}
}