《剑指offer》学习笔记_面试题16_数值的整数次方

#include <cfloat>

class Solution {
public:
    double Power(double base, int exponent) {
        //0的负数次方直接返回
        if(equal(base,0.0)&&exponent<0)return 0.0;
        unsigned int absExponent = abs(exponent);
        double res = PowerWithUnsignedExponent(base,absExponent);
        if(exponent<0)res = 1.0/res;//负数次方
        return res;
    }
    //判断两个double类型的值是否相等
    bool equal(double i,double j){
        int diff = abs(i-j);
        if(diff<=DBL_EPSILON)return true;
        else return false;
    }
    //求正整数次幂
    double PowerWithUnsignedExponent(double base,unsigned int exponent){
        if(exponent==0)return 1;
        if(exponent==1)return base;
        //分治
        double res = PowerWithUnsignedExponent(base,exponent>>1);
        res *= res;
        //如果exponent对应二进制数末尾为1，exponent&0x1结果为1；否则exponent&0x1结果为0；
        //exponent对应二进制数的末尾为1，则exponent是奇数，否则为偶数；
        if(exponent&0x1==1)res *= base;
        return res;
    }
};