[50. Pow(x, n)](https://leetcode.com/problems/powx-n/)

### 題目描述

Implement [pow(x, n)](https://cplusplus.com/reference/valarray/pow/), which calculates `x` raised to the power `n` (i.e., x^n^).

### 範例

**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
* -2^31^ <= `n` <= 2^31^-1
* `n` is an integer.
* Either `x` is not zero or `n > 0`.
* -10^4^ <= x^n^ <= 10^4^

### 解答
#### Javascript

function myPow(x, n) {
  let result = 1;
  while (n) {
    if (n % 2) {
      if (n > 0) {
        result *= x;
      } else {
        result /= x;
      }
    }

    x = x * x;
    n = ~~(n / 2);
  }

  return result;
}

> [name=Marsgoat][time=Jul 24, 2023]

#### C++

class Solution {
public:
    double myPow(double base, int n) {
        long long exponent = (long long) n;
        if (exponent == 0) {
            return 1;
        }
        if (exponent < 0) {
            base = 1 / base;
            exponent = -exponent;
        }
        double ans = 1;
        while (exponent > 0) {
            if (exponent % 2 == 1) {
                ans *= base;
                exponent -= 1;
            }
            base = base * base;
            exponent /= 2;
        }
        return ans;
    }

    template<typename T>
    void printVector(const vector<T>& vec) {
        for (const T& element : vec) {
            cout << element << ",";
        }
        cout << endl;
    }
};

> [name=Jerry Wu][time=24 July, 2023]

### Reference

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