剑指offer--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。

public class Solution {    public double Power(double base, int exponent) {        if(exponent == 0)            return 1;        int flag = 0;        double temp = 1;        double sum = base;        if(exponent > 0)            flag = 1;        else            exponent = -exponent;        while(exponent>1){            if(exponent%2!=0){                temp = temp*sum;            }            sum = sum*sum;            exponent = exponent/2;        }        if(flag == 1)            return sum*temp;        else            return 1/(sum*temp);    }}