# 50. Pow(x, n)

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 [−2^31, 2^31 − 1]

# Solution

Approach 1: Recursion.

Approach 2: Iteration.

# Code (Python)

Approach 1:

class Solution:
    def myPow(self, x: float, n: int) -> float:
        if n == 0:
            return 1
        if n == 1:
            return x
        if n == -1:
            return 1 / x
        if n % 2 == 1:
            half = self.myPow(x, n // 2)
            return half * half * x
        else:
            half = self.myPow(x, n // 2)
            return half * half

Approach 2:

# Code (C++)

Approach 1:

// Recursion.
// Growth: 1, 2, 4, 8, ...
class Solution {
public:
    double myPow(double x, int n) {
        if (n == 0) return 1;
        if (n < 0) x = 1 / x;
        long absLongN = abs(long(n));
        double pow = myPow(x, absLongN / 2);
//        if (n % 2 == 0)
        if (!(n & 1))
            pow = pow * pow;
        else
            pow = pow * pow * x;
        return pow;
    }
};

Approach 2:

// Iteration.
// Growth: 1, 2, 3, 4, ... [..., 4, 3, 2, 1]
class Solution {
public:
    double myPow(double x, int n) {
        long absLongN = abs(long(n));
        double pow = 1;
        int growthRate = 1;
        double growthBase = x;
        while (absLongN > 0)
        {
            if (absLongN >= growthRate)
            {
                absLongN -= growthRate;
                pow *= growthBase;
                growthRate++;
                growthBase *= x;
            }
            else
            {
                growthRate--;
                growthBase /= x;
            }
        }
        if (n < 0)
            pow = (double)1 / pow;
        return pow;
    }
};