Oct 5, 2021 - 3 min ' read

Pow(x, n)

Tags : recursion, leetcode, cpp, easy

Problem Statement - link #

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

Examples #

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
-10^4 <= x^n <= 10^4

Solution #

class Solution {
public:
    double myPow(double x, int nx) {
        long n = nx;
        if( n == 0 ){
            return 1;
        }
        if( n < 0 ){
            n = -n;
            x = 1/x;
        }
        if( n == 1){
            return x;
        }
        if( n%2 == 0 ){
            return myPow(x*x,n/2);
        }else{
            return x*myPow(x*x,(n-1)/2);
        }

    }
};