Java算法求水仙花数

import java.math.BigInteger;import java.util.Hashtable;public class Main {private static final int SIZE = 21;private int[] countArray = new int[10]; // 个数列表private int[] countSumArray = new int[10]; // 个数总数private BigInteger[] sumArray = new BigInteger[10];// 值总数private int offset = 0;// 浮标/*** 设置当前浮标对应的个数，个数的总数，值总数** @param num* 个数*/private void setValue(int num) {countArray[offset] = num;if (offset == 0) {countSumArray[offset] = num;sumArray[offset] = p(9 - offset).multiply(n(num));} else {countSumArray[offset] = countSumArray[offset - 1] + num;sumArray[offset] = sumArray[offset - 1].add(p(9 - offset).multiply(n(num)));}}/*** 检验当前数据是否匹配** @return*/private boolean checkPersentArray() {BigInteger minVal = sumArray[offset];// 当前已存在值BigInteger maxVal = sumArray[offset].add(p(9 - offset).multiply(n(SIZE - countSumArray[offset])));// 当前已存在值+可能存在的最大值// 最小值匹配if (pareTo(MAX) > 0) {return false;}// 最大值匹配if (pareTo(MIN) < 0) {return false;}String minStr = pareTo(MIN) > 0 ? minVal.toString() : MIN.toString();String maxStr = pareTo(MAX) < 0 ? maxVal.toString() : MAX.toString();// 找到最小值与最大值间首部相同的部分int[] sameCountArray = new int[10];for (int i = 0; i < SIZE; i++) {char c;if ((c = minStr.charAt(i)) == maxStr.charAt(i)) {sameCountArray[c - '0'] = sameCountArray[c - '0'] + 1;} else {break;}}// 判断如果相同部分有数据大于现在已记录的位数，返回falsefor (int i = 0; i <= offset; i++) {if (countArray[i] < sameCountArray[9 - i]) {return false;}}// 如果当前值的总数为SIZE位，那么判断该值是不是需要查找的值if (countSumArray[offset] == SIZE) {String sumStr = sumArray[offset].toString();BigInteger sum = ZERO;for (int i = 0; i < sumStr.length(); i++) {sum = sum.add(p(sumStr.charAt(i) - '0'));}return pareTo(sumArray[offset]) == 0;}return true;}/*** 退出循环，打印** @return*/private void success() {System.out.println("find a match number:" + sumArray[offset]);}/*** 将浮标指向下一位数** @return*/private void next() {offset++;setValue(SIZE - countSumArray[offset - 1]);}/**** 回退浮标，找到最近的浮标，并减一** @return*/private boolean back() {// 回退浮标，找到最近的浮标，并减一if (countArray[offset] == 0) {while (countArray[offset] == 0) {if (offset > 0) {offset--;} else {return true;}}}if (offset > 0) {setValue(countArray[offset] - 1);return false;} else {return true;}}/*** 测试程序** @param startValue* 测试匹配数中包含9的个数* @param startTime* 程序启动时间*/private void test(int startValue, long startTime) { // 设置9的个数offset = 0;setValue(startValue);while (true) {if (checkPersentArray()) {// 检查当前提交数据是否匹配// 匹配且总数正好为SIZE的位数，那么就是求解的值if (countSumArray[offset] == SIZE) {success();}// 总数不为SIZE，且当前值不在第10位（即不等于0）if (offset != 9) {next();continue;}// 总数不为SIZE，且当前值在第10位。

if (back()) {break;}} else {if (back()) {break;}}}System.out.println(Thread.currentThread() + " End,Spend time " + (System.currentTimeMillis() - startTime) / 1000 + "s");}/*** 主函数*/public static void main(String[] args) {final long startTime = System.currentTimeMillis();int s = SIZE > 9 ? 9 : SIZE;for (int i = 0; i <= s; i++) {// new Main().test(i, startTime);// 启动十个线程同时运算final int startValue = i;new Thread(new Runnable() {public void run() {new Main().test(startValue, startTime);}}).start();}}private static final BigInteger ZERO = new BigInteger("0");private static final BigInteger MIN;private static final BigInteger MAX;/*** 0-SIZE间的BigInteger*/private static final BigInteger n(int i) {return (BigInteger) ht.get("n_" + i);}/*** 0-9的次方的BigInteger*/private static final BigInteger p(int i) {return (BigInteger) ht.get("p_" + i);}/*** 用于缓存BigInteger数据，并初始化0-SIZE间的BigInteger和0-9的次方的BigInteger*/private static Hashtable<String, Object> ht = new Hashtable<String, Object>();static {int s = SIZE < 10 ? 10 : SIZE;for (int i = 0; i <= s; i++) {ht.put("n_" + i, new BigInteger(String.valueOf(i)));}for (int i = 0; i <= 10; i++) {ht.put("p_" + i, new BigInteger(String.valueOf(i)).pow(SIZE));}MIN = n(10).pow(SIZE - 1);MAX = n(10).pow(SIZE).subtract(n(1));}}。