求解21位花朵数的解法

import java.math.BigInteger;import java.util.Arrays;/** * 如题，求解21位花朵数的解法，要求在3分钟之内计算完成花朵数： N位整数，它等于各个位的数字的N次方之和， * * 例如有一个N位数字，a1a2a3a4.....aN = a1^N +a2^N+......aN^N * * @author hb */public class FlowerNum {private static int num = 21;private static BigInteger[] table = new BigInteger[10];private static int[] nums;public static void main(String[] args) {long time = System.currentTimeMillis();for (int i = 0; i < 10; i++)table[i] = BigInteger.valueOf(i).pow(num);nums = new int[num];find(nums, 0, 0);time = System.currentTimeMillis() - time;System.out.println(time / 1000.0 + "s");}/** * level表示第几位数，num表示第几位数的数字； */public static void find(int[] nums, int level, int num) {nums[level] = num;if (level == nums.length - 1) {BigInteger big = sum(nums);//当level为20时，表示nums数组下标为20即21位数字满后求和；int[] temp = getArray(big);//将得到的21位数和转化为整形数组；并返回；if (check(nums, temp))//测试是否是21位，并将数组复制给nums数组；System.out.println(big);//输出big数；return;}for (int i = num; i < 10; i++)//递归；查找21位数字符合条件的数；find(nums, level + 1, i);}/** * 检测长度，苦符合条件就复制a2到a1； * @param a1 * @param a2 * @return */public static boolean check(int[] a1, int[] a2) {if (a1.length != a2.length)return false;Arrays.sort(Arrays.copyOf(a1, a1.length));Arrays.sort(a2);return Arrays.equals(a1, a2);}/** * 得到nums数组的数做为table下标的数的和。并返回； * @param nums * @return */public static BigInteger sum(int[] nums) {BigInteger sum = BigInteger.ZERO;for (int i = 0; i < nums.length; i++)sum = sum.add(table[nums[i]]);return sum;}/** * 将BigInteger转化为int[]类型数据； * @param big * @return */public static int[] getArray(BigInteger big) {String s = String.valueOf(big);char[] ch = s.toCharArray();int[] res = new int[ch.length];for (int i = 0; i < ch.length; i++)res[i] = ch[i] - '0';return res;}}