428. Pow(x, n)medium

Posted 2020-10-23 abc_begin

Implement pow(x, n).

 Notice

You don‘t need to care about the precision of your answer, it‘s acceptable if the expected answer and your answer ‘s difference is smaller than 1e-3.

 

Example

Pow(2.1, 3) = 9.261
Pow(0, 1) = 0
Pow(1, 0) = 1

Challenge 

O(logn) time

参考了@grandyang 的代码

 

解法一：

 1 class Solution {
 2 public:
 3     double myPow(double x, int n) {
 4         double res = 1.0;
 5         for (int i = n; i != 0; i /= 2) {
 6             if (i % 2 != 0) {
 7                 res *= x;
 8             }
 9             x *= x;
10         }
11         return n < 0 ? 1 / res : res;
12     }
13 };        

迭代

 

解法二：

 1 class Solution {
 2 public:
 3     double myPow(double x, int n) {
 4         if (n < 0) {
 5             return 1 / power(x, -n);
 6         }
 7 
 8         return power(x, n);
 9     }
10     double power(double x, int n) {
11         if (n == 0) {
12             return 1;
13         }
14 
15         double half = power(x, n / 2);
16         if (n % 2 == 0) {
17             return half * half;
18         }
19         
20         return x * half * half;
21     }
22 };

 

解法三：

 1 class Solution {
 2 public:
 3     /**
 4      * @param x the base number
 5      * @param n the power number
 6      * @return the result
 7      */
 8     double myPow(double x, int n) {
 9         if (n == 0) {
10             return 0;
11         }
12         if (n == 1) {
13             return x;
14         }
15         if (n == -1) {
16             return 1 / x;
17         }
18 
19         return myPow(x, n / 2) * myPow(x, n - n / 2);
20     }
21 };

会超时