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

 50Pow(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;
    }
};


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