Implement pow(x, n).
Notice
You don't need to care about the precision of your answer, it's acceptable if the expected answer and your answer 's difference is smaller than 1e-3.
Example
Pow(2.1, 3) = 9.261
Pow(0, 1) = 0
Pow(1, 0) = 1LeetCode上的原题，请参见我之前的博客Pow(x, n)。
解法一：
class Solution { public: /** * @param x the base number * @param n the power number * @return the result */ double myPow(double x, int n) { if (n == 0) return 1; double half = myPow(x, n / 2); if (n % 2 == 0) return half * half; else if (n > 0) return half * half * x; else return half * half / x; } };
解法二：
class Solution { public: /** * @param x the base number * @param n the power number * @return the result */ double myPow(double x, int n) { if (n == 0) return 0; if (n == 1) return x; if (n == -1) return 1 / x; return myPow(x, n / 2) * myPow(x, n - n / 2); } };
本文转自博客园Grandyang的博客，原文链接：求x的n次方[LintCode] Pow(x, n) ，如需转载请自行联系原博主。
