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[LeetCode] 50. Pow(x, n) #102

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Animenzzzz opened this issue Sep 29, 2019 · 0 comments
Open

[LeetCode] 50. Pow(x, n) #102

Animenzzzz opened this issue Sep 29, 2019 · 0 comments

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@Animenzzzz
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题目描述:

实现 pow(x, n) ,即计算 x 的 n 次幂函数。

示例 1:

输入: 2.00000, 10
输出: 1024.00000

示例 2:

输入: 2.10000, 3
输出: 9.26100

示例 3:

输入: 2.00000, -2
输出: 0.25000
解释: 2-2 = 1/22 = 1/4 = 0.25

说明:

-100.0 < x < 100.0
n 是 32 位有符号整数,其数值范围是 [−231, 231 − 1] 。

解题思路:使用递归,每次折半来进行运算。

C++解题:

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