50. Pow(x, n) (Medium)

https://leetcode.com/problems/powx-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 [−2³¹, 2³¹ − 1]

Solutions

class Solution {
    // keep results already calculated, avoid repetitive computation
    Map<String, Double> cache = new HashMap<>();

    public double myPow(double x, int n) {
        if (n == 0) {
            return 1d;
        }

        if (x == 1) {
            return x;
        }

        // Use long type instead of int such that avoid overflow issue
        long newN = n;
        if (newN < 0) {
            x = 1 / x;
            newN = -newN;
        }

        return solve(x, newN);
    }

    public double solve(double x, long n) {
        if (n == 0) {
            return 1;
        }

        if (n == 1) {
            return x;
        }

        String key = x + "#" + n;
        if (cache.containsKey(key)) {
            return cache.get(key);
        }

        double val = solve(x, n / 2);

        // if n is even
        double ans = val * val;

        // if n is odd, do not forget there is still 1 x not multiplied.
        if (n % 2 == 1) {
            ans = val * val * x;
        }

        cache.put(key, ans);

        return ans;
    }
}

Incorrect Solutions

References