Leetcode: Pow(x, n)

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Implement pow(x, n).

Analysis:

The most basic idea, x multiply itself n time and return.  However, this strait forward approach got time excessed. O(n).

Note that O(logn)< O(n)<O(n^2)<O(2^n).  Therefore, we should consider about recursive approach, and divide n by 2 for each recursive call.

 1 public class Solution { 2     public double pow(double x, int n) { 3         if (n < 0) { 4             return 1/helper(x, -n); 5         } 6         else return helper(x, n); 7     } 8      9     public double helper(double x, int n) {10         if (n == 0) return 1;11         double half = helper(x, n/2);12         if (n%2 == 0) {13             return half*half;14         }15         else return half*half*x;16     }17 }

參考網上的想法，把x的n次方劃分成兩個x的n/2次方相乘，然後遞迴求解子問題，結束條件是n為0返回1。因為是對n進行二分，演算法複雜度和上面方法一樣，也是O(logn)。

Leetcode: Pow(x, n)