50. Pow(x, n)
题目链接:https://leetcode-cn.com/problems/powx-n/
解题思路:
分奇偶数。
1 public class Solution { 2 public double myPow(double x, int n) { 3 int sign=1; 4 if(n<0){ 5 n=-n; 6 return 1/pow(x,n); 7 } 8 9 else 10 return pow(x,n); 11 } 12 13 14 public double pow(double x,int n){ 15 if(n==0) return 1; 16 if(n==1) return x; 17 if(n%2==0) { 18 double tmp=pow(x,n/2); 19 return tmp*tmp; 20 }else{ 21 double tmp=pow(x,n/2); 22 return tmp*tmp*x; 23 } 24 } 25 26 }