Leetcode 50 POW (x, y)

Implement POW (X,N), Which calculatesXRaised to the powerN(Xn ).

Example 1:

Input: 2.00000, 10Output: 1024.00000

Example 2:

Input: 2.10000, 3Output: 9.26100

Example 3:

Input: 2.00000, -2Output: 0.25000Explanation: 2-2 = 1/22 = 1/4 = 0.25

Note:

• -100.0 <X<100.0
• NIs a 32-bit signed integer, within the range [? 231,231? 1]

Maybe I think too much, because the negative number range of the int is more than the positive number range. to convert it to a positive integer, we need to consider another range.

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;        }    }};

I still feel bad. I found an article well written.

80021207

Leetcode 50 POW (x, y)