leetcode 372. Super Pow 快速幂实现

Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array.

Example1:

a = 2
b = [3]

Result: 8
Example2:

a = 2
b = [1,0]

Result: 1024

这道题就是实现幂运算，但是不会做，网上看到了一个做法，很不错。

代码如下：

//快速幂，涉及到数学知识，不会做public class Solution{    public int superPow(int a, int[] b)    {        if(a == 1 || b.length == 0)            return 1;        a = a % 1337;        int[] nums = new int[11];        int prev = 1,i = 0;        while(i <= 10){//至于为什么是10，因为更多的话会浪费空间，更小的话，会增加递归层数，至于10是否最优，没有测量。            nums[i++] = prev;            prev = prev * a % 1337;        }        int[] divisor = new int[b.length - 1];        int remainder = divide(b,10,divisor);        return superPow(nums[10],divisor)*nums[remainder]%1337;    }    private int divide(int[] num,int baseNum,int[] divisor)    {        System.arraycopy(num,0,divisor,0,divisor.length);        return num[num.length - 1];    }}