刷题记录-剑指offer16：数值的整数次方

Posted 2020-11-29 tendermelon

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。

使用递归时间复杂度降为o（logn）n为指数大小

对cyc大佬的补充，考虑了base为零的情况，否则当base==0，exp为复数是会出错

public class Solution {
    public double Power(double base, int exponent) {
        if(Double.toString(base).equals(Double.toString(0.0)))
            return 1.0;
        if(exponent == 1)
            return base;
        boolean isNegtive = false;
        if(exponent < 0){
            exponent = -exponent;
            isNegtive = true;
        }
        double result = Power(base*base, exponent/2);
        if(exponent % 2 !=0)
            result *= base;
        return isNegtive == false? result: 1.0/result;
  }
}