LeetCode 50. Pow(x, n) (Medium)

题目

Implement pow(x, n), which calculates x raised to the power n (i.e., xn).
 
Example 1:

Input: x = 2.00000, n = 10
Output: 1024.00000

 
Example 2:

Input: x = 2.10000, n = 3
Output: 9.26100

 
Example 3:

Input: x = 2.00000, n = -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25

 
Constraints:

-100.0 < x < 100.0
-231 <= n <= 231-1
-104 <= xn <= 104

 

思路

重点在于n在奇数时，n为正数和负数两种情况的处理不一样。
当n为奇数
n为正时，return pow(x, n/2)*pow(x, n/2)*pow(x, 1)
n为负时，return pow(x, n/2)*pow(x, n/2)*pow(x, -1)

class Solution {
    public double myPow(double x, int n) {
        if(n == 0) return 1;
        double half = myPow(x, n/2);
        if((n%2) == 0) return half * half;
        if(n > 0) return half * half * x;
        return half * half / x;
    }
}