LeetCode: Pow(x, n)

Problem:

Implement pow(x, n).

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 考虑到指数为负的情况 利用二分来求结果 时间复杂度O(logn) 

class Solution {
public:
    double myPow(double x, int n) {
        if(n<0) 
            return 1.0/mypow(x,-n);
        return mypow(x,n);
    }
    
    double mypow(double base,unsigned int expo)
    {
        if(expo==0) return 1;
        
        double result=mypow(base,expo >> 1);
            result*=result;
        if(expo & 0x1 == 1)
            result*=base;
        return result;
    }
};