50. Pow(x, n)

Description of Problem

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

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

Solution

Explantion

The core of the problem is to consider all edge cases and remember to handle the overflow/underflow cases. The calculation can be done by "Divide and Conquer" approach.

Code (Rust)

impl Solution {
    fn my_pow(x : f64, n: i32) -> f64{
        match (x, n){
            (0.0, _) => 0.0,
            (-1.0, n) => if n % 2 == 1 {-1.0} else {1.0},
            (1.0, _) | (_, 0) => 1.0,
            (x, i32::MIN) => 1.0 / Self::my_pow(x, -(n+1)),
            (x, n) => {
                if n < 0 {
                    1.0 / Self::my_pow(x, -n)
                }
                else{
                    let v = Self::my_pow(x, n/2);
                    match (n % 2 == 0) {
                        true => v * v,
                        false => v * v * x,
                    }
                }
            }
        }
    }

}

Complexity

n is the exponent

Time complexity:

\(T(n)=O(\log(n))\)
- In general, the number of functional calls is bound by \(\log(n)\).

Auxiliary Space:

\(S(n)=O(\log(n))\)
- Same reason as the above-mentioned.

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