LeetCode-50-Pow(x,n)

算法描述：

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]

解题思路：二分法。

    double myPow(double x, int n) {
        if(n >= 0) 
            return genPow(x, n);
        else
            return 1.0/genPow(x, n);
    }
    
    double genPow(double x, int n){
        if(n == 0) return 1;
        double y = genPow(x, n/2);
        if(n%2==0)
            return y*y;
        else
            return y*y*x;
    }