剑指Offer(9)

题目:

  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。

解法:

  要考虑到底数为0,指数为负数的情况,这道题主要考的是对边界值的思考。

public class Solution {
    public double Power(double base, int exponent) {
        boolean g_InvaildInput = false;
        
        if(equals(base,0.0) && exponent<0){
            g_InvaildInput = true;
            return 0.0;
        }
        
        int absExponent = exponent;
        if(exponent<0)
            absExponent = -exponent;
        
        double result = PowerWithExponent(base,absExponent);
        
        if(exponent<0){
            result = 1.0 / result;
        }
        
        return result;
  }
    
    private double PowerWithExponent(double base, int exponent){
        double result = 1.0;
        for(int i = 1;i<=exponent;i++){
            result *= base;
        }
        return result;
    }
    
    private boolean equals(double numberOne,double numberTwo){
        return Math.abs(numberOne - numberTwo)<0.000001;
    }
}
原文地址:https://www.cnblogs.com/figsprite/p/10462548.html