題目描述
給定一個(gè)double類型的浮點(diǎn)數(shù)base和int類型的整數(shù)exponent构拳。求base的exponent次方棋弥。
public class Solution {
public double Power(double base, int exponent){
if(exponent == 0){
return 1;
}
if(exponent < 0) {
int temp = - exponent;
double result = recursion(base, temp);
result = 1/result;
return result;
}else {
double result = recursion(base, exponent);
return result;
}
}
private double recursion(double base, int exponent) {
if(exponent == 1) {
return base;
}
if(exponent % 2 == 0) {
double temp = recursion(base,exponent/2);
temp *= temp;
return temp;
}else {
double temp = recursion(base,exponent/2);
temp *= temp * base;
return temp;
}
}
}
