[LeetCode74]Pow(x,n)

Implement pow(x, n).

幂运算 Calculates x raised to the power of y.

Analysis:

If we just use the normal way of calculation, when face 1 to the power of 10000, the computation complexity is too high.

Consider this way:
If we want to get 2^10,

2^10 = 2^4 * 2^4 *2^2
2^4 = 2^2*2^2
2^2 = 2*2

Let's see this in a bottom-up way, totally it needs to loop n/2 > 0 times, first we have 2^1=2, then we can have 2^2 = 2*2=4, and again we have 2^4 = 4*4, (key point:) and if n%2==1, we need one more previous value, which is 2^2 here, so we have 2^4*2^4*2^2 = 2^10.

We use binary divide method.

Java

public double pow(double x, int n) {
		if(n<0){
			return 1.0/power(x, -n);
		}else {
			return power(x, n);
		}
    }
	public double power(double x, int n){
		if(n==0)
			return 1;
		double v = power(x, n/2);
		if(n%2==0)
			return v*v;
		else
			return v*v*x;
	}

c++

double power(double x, int n){
    if(n==0)
        return 1;
    double v = power(x,n/2);
    if(n%2 == 0)
        return v *v;
    else
        return v* v* x;
}
double pow(double x, int n) {
    if(n<0)
        return 1.0 / power(x,-n);
    else
        return power(x,n);
}