LeetCode OJ-- Pow(x, n) **@

https://oj.leetcode.com/problems/powx-n/

二分法 

class Solution {
public:
    double pow(double x, int n) {
        double ans = 1;
        
        if(n == 0)
            return 1.0;
        if(n == 1)
            return x;    
        bool isNegative = false;
        if(n < 0)
        {
            isNegative = true;
            n = -1 * n;
        }
        double n1 = n / 2;
        double n2 = n - n1*2;
        double t1 = pow(x,n1); //recursively
        double t2 = pow(x,n2); //recursilvely
        ans = t1*t1*t2;
        if(isNegative)
            ans = 1.0/ans;
            
        return ans;
    }
};