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
class Solution {
public double myPow(double x, int n) {
if(n == 0)
return 1;
if(n<0){
// 区分分数场景
return 1/x * myPow(1/x, -(n + 1));
}
// 区分奇数和偶数
return (n%2 == 0) ? myPow(x*x, n/2) : x*myPow(x*x, n/2);
}
}