[Leetcode] Pow (x, n)

Pow (x, n)

Implement POW (X,N).

Algorithm ideas:

There is nothing to say about the two-way question. Just be careful. This question is time-consuming.

Return POW (x, N> 1) * POW (x, N> 1) cannot pass. Therefore, POW (x, n/2) is obtained first. In fact, the time complexity is the same.

[Note]: The value range of N; n = integer. min_value,-N is out of range, so I used long for packaging.

The Code is as follows:

 1 public class Solution { 2   public double pow(double x, int n) { 3         return pow(x,(long)n); 4     } 5     public double pow(double x,long n){ 6         if(x == 0 || x == 1) return x; 7         if(n == 0) return 1; 8         if(n == 1) return x; 9         if(n < 0) return 1 / pow(x,-n);10         double tem = pow(x,n >> 1);11         if((n & 1 )== 0){12             return tem * tem;        13         }else{14             return tem * tem * x;                    15         }16     }17 }