leetcode 二分法 Pow(x, n)

Posted 2020-09-26 cxchanpin

Pow(x, n)

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Implement pow(x, n).

题意：求x的n次幂
思路：二分法
n有可能是负的或正的
当n为负是，pow(x, n) = 1/pow(x, -n)
x^n = x^{n/2} * x^{n/2}* x^{n%2}
复杂度：时间O(log n)。空间O(1)

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 power(x, n);
	}else{
		return 1/power(x, -n);
	}
}