LeetCode 50: Pow(x, n)

最新推荐文章于 2025-04-08 15:46:54 发布

claviclelau

最新推荐文章于 2025-04-08 15:46:54 发布

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本文提供了两种实现幂运算的方法：一种是递归方式，考虑了多种边界情况；另一种是非递归方式，通过循环逐步计算幂的结果。文章重点介绍了每种方法的细节，并对特殊情况进行了解释。

Difficulty: 3

Frequency: 5

Problem:

Implement pow(x, n).

Solution:

1. Recursive solution from AnnieKim. Lots of subtleties.

class Solution {
public:
    double pow(double x, int n) {
        if (x < 0)
            return (n % 2 == 0) ? pow(-x, n) : -pow(-x, n);
        if (x == 1)
            return x;
        if (n < 0)
            return 1.0 / pow(x, -n);
        if (n == 0)
            return 1.0;
        if (n == 1)
            return x;
        double half = pow(x, n / 2);
        if (n % 2 == 0)
            return half * half;
        else
            return x * half * half;
    }
};

2. Not that recursive one from MoreWindows.

class Solution {
public:
    double powNotThatRecursive(double x, int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (x==1||n==0)
            return 1;
            
        if (n==1)
            return x;
        
        if (x<0)
            return (n&1)==0?pow(-x, n):-pow(-x, n);
            
        if (n<0)
            return 1.0/pow(x, -n);
           
        double answer = 1;
        while(n!=0)
        {
            if ((n&1)!=0)
                answer *= x;
            
            x *= x;
            n >>= 1;
        }
        return answer;
    }
};

Notes:

Thanks to AnnieKim and MoreWindows. This problem is very important. Should pay special attention to it.