50. Pow(x, n)

50. Pow(x, n) (Medium)

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]

Related Topics:
Math, Binary Search

Similar Questions:

• Sqrt(x) (Easy)
• Super Pow (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/powx-n/
// Author: github.com/lzl124631x
// Time: O(logN)
// Space: O(logN)
class Solution {
    double myPow(double x, long n) {
        if (n < 0) return 1 / myPow(x, -n);
        if (n == 0) return 1;
        if (n == 1) return x;
        if (n == 2) return x * x;
        return myPow(myPow(x, n / 2), 2) * (n % 2 ? x : 1);
    }
public:
    double myPow(double x, int n) {
        return myPow(x, (long)n);
    }
};