[LeetCode]Pow(x, n)

最新推荐文章于 2025-04-08 15:46:54 发布

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本文介绍了一种使用C++实现快速幂运算的方法，通过递归分治的方式提高计算效率，适用于整数和浮点数的幂次方计算。

class Solution {
//divide-and-conquer
//classic
public:
	double pow(double x, int n) {
		if (n == 0) return 1.0;
		// Compute x^{n/2} and store the result into a temporary
		// variable to avoid unnecessary computing
		double half = pow(x, n / 2);
		if (n % 2 == 0)
			return half * half;
		else if (n > 0)
			return half * half * x;
		else
			return half * half / x;
	}
};

second time

class Solution {
public:
    double pow(double x, int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(n == 0) return 1;
        else if(n > 0)
        {
            double half = pow(x, n/2);
            if(n%2 == 0) return half*half;
            else return half*half*x;
        } 
        else
        {
            n = -n;
            double half = pow(x, n/2);
            if(n%2 == 0) return 1.0/(half*half);
            else return 1.0/(half*half*x);
        }
    }
};