let 50 Pow(x,n)

快速幂算法

public double quickPow(double x,int n){
if(n==0) return 1.0000;
double ans=1.0;
while(n!=0){

if(n%2!=0){
ans=ans*x;
}
x*=x;
n/=2;
}
return ans;
}

class Solution {
public double myPow(double x, int n) {

boolean divide= n<0? true: false;
n=Math.abs(n);
double ans=quickPow(x,n);
if(divide)
ans=1.0/ans;
return ans;
}

public double quickPow(double x,int n){
if(n==0) return 1.0000;
double ans=1.0;
while(n!=0){

if(n%2!=0){
ans=ans*x;
}
x*=x;
n/=2;
}
return ans;
}
}