Implement pow (x, n) function

This question cannot be used for Loop and multiply , Will timeout . This problem uses the fast power method .
  Using binary and decimal conversion method , hold n Open .

In a nutshell ,n One right at a time ,x Multiply yourself every time , When n The rightmost of is 1 when ,res = res * x;

 Such as x^10^ Calculate with the above method ：
	10 The binary of is 1010    Records of the results res = 1
	1、1010 On the right is 0, So do not calculate , then 1010 Shift right to 101,x = x*x, Turn into x The square of 
	2、101 On the far right is 1, Calculation  res = res * x;  namely res = x The square of ,101 Move right to 10,x = x*x , here x For the initial x Of 4 Power 
	3、10 On the far right is 0, Don't count , Shift right to 1,x = x*x;x Of the initial value 8 Power 
	4、1 On the far right is 1, Calculation res = res * x; namely res = x The square of  * x Of 8 Power , 1 Move right to 0, x = x * x
	5、 end 
 From the above steps , Is to put x Of 10 To the power of x Of 8 The power and x Square multiplication of 
 Update every cycle x The value of is x The square of ,x In turn 1 Power 2 Power 4 Power 8 Power 16 Power ..... When n On the far right is 1 when , Then record this time and x The product of the . 

  public double myPow(double x, int n) {
    
        if(x==0) return 0;
        long b = n;
        double res = 1;
        if(b<0){
    
            x = 1/x;
            b = -b;
        }
        while(b>0){
    
            if((b&1)==1) res *= x;
            x *= x;
            b >>= 1;
        }
        return res;
    }

Java In the code int32 Variable n \in [-2147483648, 2147483647]n∈[−2147483648,2147483647] , So when n = -2147483648 n=−2147483648 When the n = -n ,n=−n The assignment will be wrong due to out of bounds . The solution is to put n Deposit in long Variable b , Later use b Just operate .