[LeetCode] 50. Pow(x, n) 求x的n次方

Implement pow(x, n).

求x的n次幂，如果用n个循环相乘，那就太简单了，肯定不是这题要考察的。

解法：利用性质：

n是奇数，n % 2 == 1, x^n = x^(n/2) * x^(n/2) * x;
n是偶数，n % 2 == 0, x^n = x^(n/2) * x^(n/2);

每次n/2，递归到下一层，直到n=0是返回1。 如果n为负数，利用x^-n = 1/x^n。

Java:

class Solution {
    public double myPow(double x, int n) {  
        if(n < 0)  
            return 1 / pow(x, -n);  
        else  
            return pow(x, n);  
    }

    public double pow(double x, int n){  
        if(n == 0)  return 1;  
        double mid=pow(x, n / 2);  
        if((n & 1) == 0)  
            return mid * mid;  
        else  
            return mid * mid * x;  
    }  
}

Python: Recursion, T: O(logn), S: O(logn)

class Solution(object):
    def myPow(self, x, n):
        if n < 0 and n != -n:
            return 1.0 / self.myPow(x, -n)
        if n == 0:
            return 1
        v = self.myPow(x, n / 2)
        if n % 2 == 0:
            return v * v
        else:
            return v * v * x

Python: Iteration, T: O(logn), S: O(1)

class Solution(object):
    def myPow(self, x, n):
        """
        :type x: float
        :type n: int
        :rtype: float
        """
        result = 1
        abs_n = abs(n)
        while abs_n:
            if abs_n & 1:
                result *= x
            abs_n >>= 1
            x *= x

        return 1 / result if n < 0 else result

C++:

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

　

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