Pow(x, n)

Implement pow(x, n).题目在这里，二分解法如下：

class Solution {
public:
    double pow(double x, int n) {
        double ans;
        if(n < 0){
            ans = power(x, -n);
            return (double)1 / ans;
        }else{
            return power(x, n);
        }
    }
    double power(double x, int n) {
        double ans;
        if(n == 0) ans=1;
        else
        {  ans=power(x*x, n/2);
           if(n%2==1) ans*=x;
         }
         return ans;
    }
};

      还可以这么写：

class Solution {
public:
    double pow(double x, int n) {
        double ans;
        if(n < 0)
            return (double)1 / power(x, -n);
        else return power(x, n);
    }
    double power(double a, int n) {
        double ans = 1;
        while(n > 0){
             if(n&1) ans *= a;   
             a *= a;
             n >>= 1;        
         }
         return ans;
    }
};

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