leetcode-50 Pow(x, n) Python Java

https://leetcode-cn.com/problems/powx-n/solution/powx-n-by-leetcode-solution/

方法一：递归
Python

class Solution:
    def myPow(self, x: float, n: int) -> float:
        def quickMul(N):
            if N == 0:
                return 1.0
            y = quickMul(N // 2)
            return y * y if N % 2 == 0 else y * y * x
        return quickMul(n) if n >= 0 else 1.0/quickMul(-n)

Java

class Solution {
    public double quickMul(double x, int N) {
        if (N == 0) {
            return 1.0;
        }
        double y = quickMul(x, N / 2);
        return N % 2 == 0 ? y * y : y * y * x;
    }
    public double myPow(double x, int n) {
        return n >= 0 ? quickMul(x, n) : 1.0 / quickMul(x, -n);
    }
}

方法二：迭代
Python

class Solution:
    def myPow(self, x: float, n: int) -> float:
        def quickMul(N):
            ans = 1.0
            x_contribute = x
            while N > 0:
                if N % 2 == 1:
                    ans *= x_contribute
                x_contribute *= x_contribute
                N //= 2
            return ans
        
        return quickMul(n) if n >= 0 else 1.0/quickMul(-n)

Java

class Solution {
    public double quickMul(double x, long N) {
        double ans = 1.0;
        double x_contribute = x;
        while (N > 0) {
            if (N % 2 == 1) {
                ans *= x_contribute;
            }
            x_contribute *= x_contribute;
            N /= 2;
        }
        return ans;
    }

    public double myPow(double x, int n) {
        long N = n;
        return N >= 0 ? quickMul(x, N) : 1.0/quickMul(x, -N);
    }
}