标签：style blog io ar color sp on div log
Implement pow(x, n).
Analysis:
x^n = x^(n/2)*x^(n/2) (*x, if n is odd).
NOTE: We need consider n<0, AND if n=Integer.MIN_VALUE, -n is actually larger than Integer.MAX_VALUE by 1, so we cannot simply take -n.
Solution:
1 public class Solution { 2 public double pow(double x, int n) { 3 if (x==0) return 0; 4 if (n==0) return 1; 5 6 double res; 7 if (n>0) 8 res = powRecur(x,n); 9 else{ 10 if (n>Integer.MIN_VALUE) 11 res = powRecur(x,-n); 12 else { 13 res = powRecur(x,-(n+1)); 14 res = res*x; 15 } 16 res = 1 / res; 17 } 18 return res; 19 20 } 21 22 public double powRecur(double x, int n){ 23 if (n==1) return x; 24 25 double temp = powRecur(x,n/2); 26 double res = temp*temp; 27 if (n%2==1) res = res*x; 28 return res; 29 } 30 }
Solution 2:
Another method is to directly take care of the negative power in the recursion function.
1 public class Solution { 2 public double pow(double x, int n) { 3 if (x==0) return 0; 4 if (n==0) return 1; 5 6 double res; 7 res = powRecur(x,n); 8 return res; 9 10 } 11 12 public double powRecur(double x, int n){ 13 if (n==1) return x; 14 if (n==-1) return 1/x; 15 16 double temp = powRecur(x,n/2); 17 double res = temp*temp; 18 if (n%2==1) res = res*x; 19 if (n%2==-1) res = res*(1/x); 20 return res; 21 } 22 }
标签：style blog io ar color sp on div log
原文地址：http://www.cnblogs.com/lishiblog/p/4125525.html
