LeetCode之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递归实现快速幂运算的方法，该方法采用分治策略，通过计算x^(n/2)来减少计算量，并能有效处理正负指数的情况。

Implement pow(x, n).

Recursive solution is easier to understand. It uses the divide-and-conquer approach, which also runs in $\Theta(\lg n)$ time. The formula is shown below:

$x^n=\begin{cases}1.0&\text{if } n=0,\\ x^{n/2}\cdot x^{n/2}&\text{if }n\ne0\text{ and }n\text{ is even,}\\ x^{\lfloor n/2\rfloor}\cdot x^{\lfloor n/2\rfloor}\cdot x&\text{if }n>0\text{ and }n\text{ is odd,}\\ x^{\lceil n/2\rceil}\cdot x^{\lceil n/2\rceil}\cdot x^{-1}& \text{if }n<0\text{ and }n\text{ is odd.}\end{cases}$

class Solution {
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;
    }
};