50. Pow(x, n)

https://leetcode.com/problems/powx-n/

Problem

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

Example 1:

Input: x = 2.00000, n = 10
Output: 1024.00000

Example 2:

Input: x = 2.10000, n = 3
Output: 9.26100

Example 3:

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

Constraints:

• -100.0 < x < 100.0

• -231 <= n <= 231-1

• -104 <= xn <= 104

Solution

class Solution {
public:
    double myPow(double x, int n) {
        if (x == 0) return 0;
        if (n < 0) return 1 / myPositivePow(x, n);
        return myPositivePow(x, n);
    }
    inline double myPositivePow(double x, int n) {
        if (n == 0) return 1;
        double pow = myPositivePow(x, n / 2);
        pow *= pow;
        if (n % 2 != 0) pow *= x;
        return pow;
    }
};

• #math

• #recursive