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Implement pow(x, n).
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.
Pow(2.1, 3) = 9.261
Pow(0, 1) = 0
Pow(1, 0) = 1
O(logn) time
public class Solution { /** * @param x the base number * @param n the power number * @return the result */ public double myPow(double x, int n) { // Write your code here if(n == 0) return 1; if(n == 1) return x; if(n > 1){ if(n % 2 == 0){ double temp = myPow(x, n / 2); return temp * temp; } else{ return myPow(x, n - 1) * x; } } else{ return 1 / myPow(x, -n); } } }
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原文地址：http://www.cnblogs.com/goblinengineer/p/5352027.html
