数值的整数次方

题目描述

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。

解题思路

• xⁿ = (x * x)^n/2 n%2 = 0;
• xⁿ = x * (x²)^n/2 n%2 = 1;

代码

public class Solution {
    public double Power(double base, int exponent) {
        if (exponent == 0){
            return 1;
        }
        if (exponent == 1){
            return base;
        }
        //当为负数时结果为其倒数
        boolean isNegative = false;
        if (exponent < 0) {
            exponent = -exponent;
            isNegative = true;
        }
        double pow = Power(base * base, exponent / 2);
        if (exponent % 2 != 0){
            pow = pow * base;
        }
        return isNegative ? 1 / pow : pow;
  }
}