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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; } };
The text was updated successfully, but these errors were encountered:
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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:
示例 2:
示例 3:
说明:
解题思路:使用递归,每次折半来进行运算。
C++解题:
The text was updated successfully, but these errors were encountered: