How to implement Pow (xQuery n) by LeetCode

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1. Brief introduction of the problem.

Realize pow (x, n), that is, calculate the n-th power function of x.

2, example

Example 1:

Input: 2.00000, 10 output: 1024.00000 example 2:

Input: 2.10000, 3 output: 9.26100 example 3:

Input: 0,-2 output: 0.25000 explanation: 2-2 = 1max 22 = 1max 4 = 0.25description:

-100.0

< x < 100.0n 是 32 位有符号整数，其数值范围是 [−231, 231 − 1] 。 3，题解思路 快速幂的使用 4，题解程序 public class MyPowTest { public static void main(String[] args) { double x = 2.00000; int n = 10; double myPow = myPow(x, n); System.out.println("myPow = " + myPow); } public static double myPow(double x, int n) { if (x == 0) { return 0; } if (n == 0) { return 1; } if (n >

0) {return pow (x, n);} else {return pow (1 / x,-n);}}

Private static double pow (double x, int n) {if (n = = 0) {return 1;} double r = pow (x, n / 2); if ((n & 1) = = 1) {return r * r * x;} else {return r * r;}

5. Picture version of the problem solving program.

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