LeetCode: Pow(x, n)

Challenge Description

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

Two branch for computing pow(x, n) with divide and conquer approach.

When n is odd, pow(x, n) = x * pow(x, n-1)

When n is even, pow(x, n) = pow(x, n/2) * pow(x, n/2).

Note when x is equal with 1.0, we could do a optimization here.

Another corner case is when n is -2^31, -n will trigger a integer overflow.

#define EPS 1e-20
class Solution {
public:
    double myPow(double x, int n) {
        if(n == 1) return x;
        if(n == 0) return 1;
        if(abs(abs(x) - 1) < EPS) {
            if(x > 0) return x;
            else return n%2 ? x : -x;
        }
        if(n == INT_MIN) return 0.0;
        if(n < 0) return 1.0/(myPow(x, -n));
        if(n%2) return x * myPow(x, n - 1);
        else {
            double res = myPow(x, n/2);
            return res * res;
        }
    }
};

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