Implement pow(x, n), which calculates x raised to the power n (xn).
Example 1:
Input: 2.00000, 10 Output: 1024.00000
Example 2:
Input: 2.10000, 3 Output: 9.26100
Example 3:
Input: 2.00000, -2 Output: 0.25000 Explanation: 2-2 = 1/22 = 1/4 = 0.25
Note:
千万不要用Math.pow(x, n) 不然面试官会以为你以为他是傻逼
也不能用傻子办法一个一个乘，所以来用每次n/2，然后两项相乘得到暂时的结果，通过recursion得到最终解
class Solution { public double myPow(double x, int n) { if(n < 0){ return 1/power(x,-n); } else return power(x,n); } public double power(double x, int n){ if(n==0) return 1; double half = power(x, n/2); if(n % 2==0) return half * half; else return x*half*half; } }
简化版
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 half * half * x; return half * half / x; } };
原文：https://www.cnblogs.com/wentiliangkaihua/p/10354756.html