50

50. Pow(x, n) (Medium)

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
• n is an integer.
• Either x is not zero or n > 0.
• -104 <= xn <= 104

Companies: Amazon, Apple, Facebook

Related Topics:
Math, Recursion

Similar Questions:

• Sqrt(x) (Easy)
• Super Pow (Medium)
• Count Collisions of Monkeys on a Polygon (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);
    }
};

Solution 2.

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