50. Pow(x, n) | Leetcode Solution Implement pow(x, n), which calculates x raised to the power n (i.e., xn).

 50. Pow(x, n)

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

 

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
• -231 <= n <= 231-1
• -104 <= xn <= 104

The solution to the above question is 

class Solution {

public:

    double myPow(double x, int n) {

        double ans=1LL;

        long long nn=n;

        if(nn<0)nn=-1*nn;

        while(nn){

            if(nn%2){

                ans=x*ans;

                nn--;

            }

            else{

                x=x*x;

                nn/=2;

            }

        }

        if(n<0)ans=1LL/ans;

        return ans;

    }

};