leetcode 50 pow(x,y)

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]

可能是我想多了，因为int的负数范围比正数范围多一，转化为正整数要考虑他的一个范围

class Solution {
public:
    double myPow(double x,int n) {
        long long m=n;
        if(n<0){return pow(1/x,0-m);}
        return pow(x,n);
    }
    double pow(double x,long long n) {
        if(0==n){return 1.0;}
        double a=0.0;
        a = pow(x,n/2);
        // cout<<a<<"   "<<n<<endl;
        if(n%2){
            return a*a*x;
        }else{
            return a*a;
        }
    }
};

?感觉还是不太好，找到一篇文章写的很好

https://blog.csdn.net/gao__xue/article/details/80021207