leetcode -- Pow(x, n)

Implement pow(x, n).

直接计算会超时TLE

 1 public double pow(double x, int n) {
 2         // Start typing your Java solution below
 3         // DO NOT write main() function
 4         if(n == 0)
 5             return 1;
 6         
 7         double result = 1;
 8         if(n > 0){
 9             while(n > 0){
10                 result *= x;
11                 n --;
12             }
13         } else {
14             while(n < 0){
15                 result /= x;
16                 n ++;                
17             }
18         }
19         return result;
20     }
21 
22 // TLE test case
23 0.00001, 2147483647

使用二分法来求解，O(lgn)

这里需要注意的是当n = -2147483648，0-n=-2147483648 即整数的整数范围最大为：-2147483648~2147483647

这里需要把int 提升为long

 1 public class Solution {
 2     public double pow(double x, int n) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         long tmp = n;
 6         if(tmp >= 0){
 7             return power(x, tmp);
 8         } else {
 9             return 1.0 / power(x, -tmp);
10         }
11     }
12     
13     public double power(double x, long n){
14         if(n == 0)
15             return 1;
16             
17         if(n == 1)
18             return x;
19             
20         double result = power(x, n >> 1);
21         result *= result;
22         if((n & 0x1) == 1){
23             result = result * x;
24         }
25         
26         return result;
27     }
28 }