Implement pow(x, n), which calculates x raised to the power n (xn).
Example 1:
Input: 2.00000, 10 Output: 1024.00000
Example 2:
Input: 2.10000, 3 Output: 9.26100
Example 3:
Input: 2.00000, -2 Output: 0.25000 Explanation: 2-2 = 1/22 = 1/4 = 0.25
Note:
- -100.0 < x < 100.0
- n is a 32-bit signed integer, within the range [−231, 231 − 1]
binary search + 递归，注意n < 0时的处理
time: O(logN), space: O(logN)
class Solution { public double myPow(double x, int n) { if (n < 0) return 1 / helper(x, -n); return helper(x, n); } private double helper(double x, int n) { if (n == 0) return 1; double half = helper(x, n / 2); if (n % 2 == 0) return half * half; else return half * half * x; } }