LeeCode-Pow(x, n)

Implement pow(xn).

1 double myPow(double x, int n)
2 {
3     if(n==0)
4         return 1.0;
5     if(n<0)
6         return 1.0/pow(x,-n);
7     return x*pow(x,n-1);
8 }
时间: 2024-10-08 20:04:32

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