50. Pow(x, n)

Implement pow(x, n).

 Notice

You don't need to care about the precision of your answer, it's acceptable if the expected answer and your answer 's difference is smaller than1e-3.

Have you met this question in a real interview? 

Yes

Example

Pow(2.1, 3) = 9.261
Pow(0, 1) = 0
Pow(1, 0) = 1

分析：

如果n是偶数，那么， power (x, n) = power(x,n/2) * power(x, n/2)。

如果n是奇数，那么， power (x, n) = power(x,n/2) * power(x, n/2) * x.

注意：当负数转成正数的时候，有可能会溢出。

class Solution {
    public double myPow(double x, int n) {
        if (n == 0 || x == 1) return 1;
        if (x == 0) return 0;
        if(n == Integer.MIN_VALUE){
            x = x * x;
            n = n / 2;
        }
        if (n < 0) return 1 / myPow(x, -n);
        
        double result = myPow(x, n / 2);
        if (n % 2 == 0) {
            return result * result;
        }
        return result * result * x;
    }
}

 

class Solution {
    public double myPow(double x, int n) {
        long N = n; // use long to handle Integer.MIN_VALUE
        if (N < 0) {
            x = 1 / x;
            N = -N;
        }
        
        double result = 1.0;
        while (N > 0) {
            if (N % 2 == 1) { // if N is odd
                result *= x;
            }
            x *= x; // square the base
            N /= 2; // halve the exponent
        }
        return result;
    }
}