50. Pow(x, n) - Medium

最新推荐文章于 2025-04-01 12:00:00 发布

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最新推荐文章于 2025-04-01 12:00:00 发布

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原文链接：http://www.cnblogs.com/fatttcat/p/10063593.html

本文介绍了一种使用二分搜索和递归实现幂运算的方法，即快速幂算法。该算法能够高效计算x^n，时间复杂度为O(logN)，适用于n为32位有符号整数的情况。文章通过具体示例展示了算法的正确性和效率。

Implement pow(x, n), which calculates x raised to the power n (xn).

Example 1:

Input: 2.00000, 10
Output: 1024.00000

Example 2:

Input: 2.10000, 3
Output: 9.26100

Example 3:

Input: 2.00000, -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25

Note:

-100.0 < x < 100.0
n is a 32-bit signed integer, within the range [−231, 231 − 1]

binary search + 递归，注意n < 0时的处理

time: O(logN), space: O(logN) -- need extra space to store the computation results, there are logN computations in total

class Solution {
    public double myPow(double x, int n) {
        if (n < 0)
            return 1 / helper(x, -n);
        
        return helper(x, n);
    }
    private double helper(double x, int n) {
        if (n == 0) return 1;
        
        double half = helper(x, n / 2);
        if (n % 2 == 0)
            return half * half;
        else
            return half * half * x;
    }
}