50. Pow(x, n)

Problem Description:

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

Example 1:

Input: 2.00000, 10
Output: 1024.00000

Example 2:

Input: 2.10000, 3
Output: 9.26100

Example 3:

Input: 2.00000, -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25

Note:

• -100.0 < x < 100.0
• n is a 32-bit signed integer, within the range [−231, 231 − 1]

Code:

class Solution {
    public double myPow(double x, int n) {
        if(x == 0 && n == 0)
            return 1;
        if(x == 0) 
            return 0;
        if(n == 0 || x - 1 == 0) 
            return 1;
        // watch out the int/long about the precision issue
        double res = func(x, Math.abs((long)n));
        if(n < 0) {
            return 1/res;
        }
        return res;
    }
    double func(double x, long n) {
        if(n/2 == 0) {
            return n%2==1?x:1;    
        } 
        // pay attention to the Math.pow function
        double t;
        return n%2==1?x*(t=func(x,n/2))*t:(t=func(x,n/2))*t;
    }
}