Java实现21位水仙花数算法

1.import java.math.BigInteger;   
2.import java.util.Hashtable;   
3.  
4.public class Main {   
5.  
6.    private static final int SIZE = 21;   
7.    private int[] countArray = new int[10]; // 个数列表   
8.    private int[] countSumArray = new int[10]; // 个数总数   
9.    private BigInteger[] sumArray = new BigInteger[10];// 值总数   
10.    private int offset = 0;// 浮标   
11.  
12.    /**  
13.     * 设置当前浮标对应的个数，个数的总数，值总数  
14.     *  
15.     * @param num  
16.     *            个数  
17.     */  
18.    private void setValue(int num) {   
19.        countArray[offset] = num;   
20.        if (offset == 0) {   
21.            countSumArray[offset] = num;   
22.            sumArray[offset] = p(9 - offset).multiply(n(num));   
23.        } else {   
24.            countSumArray[offset] = countSumArray[offset - 1] + num;   
25.            sumArray[offset] = sumArray[offset - 1].add(p(9 - offset).multiply(n(num)));   
26.        }   
27.    }   
28.  
29.    /**  
30.     * 检验当前数据是否匹配  

31.     *  
32.     * @return  
33.     */  
34.    private boolean checkPersentArray() {   
35.        BigInteger minVal = sumArray[offset];// 当前已存在值   
36.        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   
37.        // 最小值匹配   
38.        if (minVal.compareTo(MAX) > 0) {   
39.            return false;   
40.        }   
41.        // 最大值匹配   
42.        if (maxVal.compareTo(MIN) < 0) {   
43.            return false;   
44.        }   
45.        String minStr = minVal.compareTo(MIN) > 0 ? minVal.toString() : MIN.toString();   
46.        String maxStr = maxVal.compareTo(MAX) < 0 ? maxVal.toString() : MAX.toString();   
47.        // 找到最小值与最大值间首部相同的部分   
48.        int[] sameCountArray = new int[10];   
49.        for (int i = 0; i < SIZE; i++) {   
50.            char c;   
51.            if ((c = minStr.charAt(i)) == maxStr.charAt(i)) {   
52.                sameCountArray[c - '0'] = sameCountArray[c - '0'] + 1;   
53.            } else {   
54.                break;   
55.            }   
56.        }   
57.        // 判断如果相同部分有数据大于现在已记录的位数，返回false   
58.        for (int i = 0; i <= offset; i++) {   
59.            if (countArray[i] < sameCountArray[9 - i]) {   
60.                return false;