Leetcode-Pow(x,n)

Implement pow(x, n).

Analysis:

x^n = x^(n/2)*x^(n/2) (*x, if n is odd).

NOTE: We need consider n<0, AND if n=Integer.MIN_VALUE, -n is actually larger than Integer.MAX_VALUE by 1, so we cannot simply take -n.

Solution:

public class Solution {
    public double pow(double x, int n) {
        if (x==0) return 0;
        if (n==0) return 1;
      
        double res;
        if (n>0)
            res = powRecur(x,n);
        else{
            if (n>Integer.MIN_VALUE)
                res = powRecur(x,-n);
            else {
                res = powRecur(x,-(n+1));
                res = res*x;
            }
            res = 1 / res;
        }
        return res;
        
    }

    public double powRecur(double x, int n){
        if (n==1) return x;

        double temp = powRecur(x,n/2);
        double res = temp*temp;
        if (n%2==1) res = res*x;
        return res;
    }
}

Solution 2:

Another method is to directly take care of the negative power in the recursion function.

public class Solution {
    public double pow(double x, int n) {
        if (x==0) return 0;
        if (n==0) return 1;
      
        double res;
        res = powRecur(x,n);
        return res;
        
    }

    public double powRecur(double x, int n){
        if (n==1) return x;
        if (n==-1) return 1/x;

        double temp = powRecur(x,n/2);
        double res = temp*temp;
        if (n%2==1) res = res*x;
        if (n%2==-1) res = res*(1/x);
        return res;
    }
}