50. Pow(x, n)

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) -- need extra space to store the computation results, there are logN computations in total

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;
    }
}

二刷：

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);
        return n % 2 == 0 ? half * half : half * half * x;
    }
}