LeetCode 题解（173): Pow(x, n)

题目：

Implement pow(x, n).

题解：

要用二分法递归。还要区分n的正负。

C++版：

class Solution {
public:
    double myPow(double x, int n) {
        if(n == 0) return 1;
        if(n == 1) return x;
        double temp = myPow(x, abs(n / 2));
        if(n > 0) {
            if(n & 1) return temp * temp * x;
            else return temp * temp;
        } else {
            if(n & 1) return 1 / (temp * temp * x);
            else return 1 / (temp * temp);
        }
    }
};

Java版：

public class Solution {
    public double myPow(double x, int n) {
        if(n == 0) return 1;
        if(n == 1) return x;
        double temp = myPow(x, Math.abs(n / 2));
        if(n > 0) {
            if((n & 1) == 0) return temp * temp;
            else return temp * temp * x;
        } else {
            if((n & 1) == 0) return 1 / (temp * temp);
            else return 1 / (temp * temp * x);
        }
    }
}

Python版：

class Solution:
    # @param {float} x
    # @param {integer} n
    # @return {float}
    def myPow(self, x, n):
        if n == 0:
            return 1.0
        if n == 1:
            return x
        if n > 1:
            return self.pow(x, n)
        else:
            return 1.0 / self.pow(x, -n)
            
    def pow(self, x, n):
        if n == 0:
            return 1.0
        if n == 1:
            return x
        temp = self.pow(x, n / 2)
        if n % 2 == 0:
            return temp * temp
        else:
            return temp * temp * x