Questions and Tests
package sword011;
/* 问题描述：实现函数double power(double base,int exponent),求base的exponent次方。不能使用库函数，同时不需要考虑大数问题。
*/
public class main {
public static void main(String[] args) {
double [] testTable = {1,2,2.1};
int [] testTable2 = {4,9,15};
for(int i=0;i<testTable.length;i++){
test(testTable[i],testTable2[i]);
}
}
private static void test(double ito,int ito2) {
Solution solution = new Solution();
double rtn;
long begin = System.currentTimeMillis();
System.out.print(ito+" ");
System.out.print(ito2);
System.out.println();
//开始时打印数组
rtn= solution.power(ito,ito2);//执行程序
long end = System.currentTimeMillis();
System.out.println("rtn=" );
System.out.print(rtn);
System.out.println();
System.out.println("耗时：" + (end - begin) + "ms");
System.out.println("-------------------");
}
}
Solution 1 (success)
First, consider the positive and negative of the exponet, and set the isPrime attribute, which is used for the final result = 1/result
. Second, a^b = a^c * a^(bc)
Using the binary method, the positive and negative of the exponet will be expanded in binary. , for example, to find 25, where 5=101(2)=1∗22+0∗21+1∗20, so 25=21∗22+0∗21+1∗20=51∗22∗50∗21∗51∗ 20
It can be seen from the above example that the positive and negative values of exponet can be written in binary. When the corresponding bit is 1, it is multiplied by the current value, otherwise it is not multiplied;
the current value is 1, 2, 4, and 8 of base. . times, powerBase*powerBase can
Refer to https://blog.csdn.net/xushiyu1996818/article/details/86607288
package sword011;
public class Solution {
public double power(double base, int exponet) {
if(exponet == 0) {
return 1;
}
if(exponet == 1) {
return base;
}
if(exponet == -1){
return 1/base;
}
if(base == 0) {
return 0;
}
// exponet是否为正
boolean isPrime = true;
if(exponet < 0) {
isPrime = false;
exponet = -exponet;
}
// 2的i次方
int power2 = 1;
// basek 的2的i次方
double powerBase = base;
double result = 1;
while(exponet != 0) {
if((exponet & power2) != 0) {
result = result * powerBase;
exponet = exponet - power2;
}
power2 = power2 << 1;
powerBase = powerBase * powerBase;
}
if(!isPrime) {
result = 1/result;
}
return result;
}
}