[Leetcode Long March series] Pow (x, n)

Original question:

Implement POW (X,N).

Idea: recursive Calculation of pow.

class Solution {public:    double pow(double x, int n) {       long long int mid = n/2;       int d = n%2;       if(n==0) return 1;       if(n==1) return x;       if(d==1) return pow(x, (n/2)+1) * pow(x, n/2);       else return pow(x, n/2) * pow(x, n/2);    }};

Timeout at the boundary value.

INT-2147483648 ~ + 2147483647 (4 bytes)

class Solution {public:    double pow(double x, int n) {       long long num = n;       if(num==0) return 1.0;       if(num==1) return x;       if(num<0) return 1.0/pow(x, -num);       double half = pow(x,num>>1);       if(num%2==0) return half*half;       else return half*half*x;    }};

If n is a negative value, the overflow occurs after negative values are obtained. Re

class Solution {public:    double pow(double x, int n) {       if(n==0) return 1.0;       if(n==1) return x;       if(n<0 && n!=INT_MIN) return 1.0/pow(x, -n);       if(n==INT_MIN) return 1.0/(pow(x, INT_MAX)*x);       double half = pow(x,n>>1);       if(n%2==0) return half*half;       else return half*half*x;    }};

Consider the minimum int_min, that is, the-2147483648 take negative will overflow, so special processing is required.

AC