分析

难度 中

来源

https://leetcode.com/problems/powx-n/

 

题目

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: 

Note:

  • -100.0 < x < 100.0
  • n is a 32-bit signed integer, within the range [−231, 231 − 1]

 

解答

Runtime: 12 ms, faster than 88.92% of Java online submissions for Pow(x, n).

 1 package LeetCode;
 2 
 3 public class L50_PowXN {
 4     public double Pow(double x, int n) {
 5         double result=1;
 6         double temp=x;
 7         String binN = Integer.toBinaryString(n);
 8         int len=binN.length();
 9         //System.out.println(binN);
10         for(int i=0;i<len;i++){
11             int pos=len-1-i;
12             //从低位到高位开始计算
13             if(binN.charAt(pos)=='1'){
14                 result*=temp;//指数位上的求和,相当于底数位上的乘积
15             }
16             temp*=temp;//当前位表示的大小
17         }
18         return result;
19     }
20     public double myPow(double x, int n) {
21         double result=0;
22         if(n==0)
23             return 1;
24         if(n<0)
25             result=1/Pow(x,-n);
26         else
27             result=Pow(x,n);
28         return result;
29     }
30     public static void main(String[] args){
31         double x=2.0;
32         int n=-2;
33         L50_PowXN l50=new L50_PowXN();
34         System.out.println(l50.myPow(x,n));
35     }
36 }
Explanation:
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