剑指offer:(11)代码的完整性： 数值的整数次方

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。

传统公式求解时间复杂度O(n）

public class Solution {    public static double Power(double base, int exponent) {        boolean g_invalid = false;        if(equal(base, 0.0) && exponent < 0){            g_invalid = true;            return 0.0;        }                 int absExponent = exponent;        if(exponent<0){            absExponent = -exponent;        }                 double result = powerWithUnsignedExponent(base,absExponent);        if(exponent<0){            result = 1.0/result;        }        return result;               }    public static boolean equal(double num1,double num2){        if(num1-num2 > -0.0000001 && num1-num2<0.0000001){            return true;        }else{            return false;        }    }    //时间复杂度为exponent-1//  public static double powerWithUnsignedExponent(double base,int exponent){//      double result = 1.0;//      for(int i = 1;i<=exponent;i++){//          result *= base;//      }//      return result;//  }    public static double powerWithUnsignedExponent(double base,int exponent){        if(exponent==0){            return 1.0;        }        if(exponent==1){            return base;        }        double result = powerWithUnsignedExponent(base, exponent>>1);            result *= result;        if((exponent & 1) == 1){            result *= base;        }        return result;    }      }