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;
}
}