面试题11：数值的整数次方

链接：https://www.nowcoder.com/questionTerminal/1a834e5e3e1a4b7ba251417554e07c00
来源：牛客网

传统公式求解时间复杂度O(n)

public class Solution {

    publicdoublePower(doublebase,int exponent) {

        double result=1;

        for(inti=0;i<Math.abs(exponent);i++){

            result*=base;

        }

        if(exponent<0){

            result=1/result;

        }

        returnresult;            

  }

}

递归：n为偶数，a^n=a^n/2*a^n/2;n为奇数，a^n=（a^（n-1）/2）*（a^（n-1/2））*a

时间复杂度O（logn） 

public class Solution {

    publicdoublePower(doublebase,int exponent) {

        intn=Math.abs(exponent);

        if(n==0)

            return1;

        if(n==1)

            returnbase;

        double result=Power(base,n>>1);

        result*=result;

        if((n&1)==1)      //若为奇数，再乘一次

            result*=base;

        if(exponent<0)

            result=1/result;

        returnresult;            

  }

}