奥古斯特顶目:Java程序如何实现对字符串的排列组合问题?

import java.math.BigInteger;
import java.util.*;

public class PermutationGenerator {

private int[] a;
private BigInteger numLeft;
private BigInteger total;
public PermutationGenerator(int n) {
if (n < 1) {
throw new IllegalArgumentException("Min 1");
}
a = new int[n];
total = getFactorial(n);
reset();
}

public void reset() {
for (int i = 0; i < a.length; i++) {
a[i] = i;
}
numLeft = new BigInteger(total.toString());
}

public BigInteger getNumLeft() {
return numLeft;
}

public BigInteger getTotal() {
return total;
}

public boolean hasMore() {
return numLeft.compareTo(BigInteger.ZERO) == 1;
}

private static BigInteger getFactorial(int n) {
BigInteger fact = BigInteger.ONE;
for (int i = n; i > 1; i--) {
fact = fact.multiply(new BigInteger(Integer.toString(i)));
}
return fact;
}

public int[] getNext() {

if (numLeft.equals(total)) {
numLeft = numLeft.subtract(BigInteger.ONE);
return a;
}

int temp;

// Find largest index j with a[j] < a[j+1]

int j = a.length - 2;
while (a[j] > a[j + 1]) {
j--;
}

// Find index k such that a[k] is smallest integer
// greater than a[j] to the right of a[j]

int k = a.length - 1;
while (a[j] > a[k]) {
k--;
}

// Interchange a[j] and a[k]

temp = a[k];
a[k] = a[j];
a[j] = temp;

// Put tail end of permutation after jth position in increasing order

int r = a.length - 1;
int s = j + 1;

while (r > s) {
temp = a[s];
a[s] = a[r];
a[r] = temp;
r--;
s++;
}

numLeft = numLeft.subtract(BigInteger.ONE);
return a;

}
//程序测试入口
public static void main(String[] args) {

int[] indices;
String[] elements = { "a", "b", "c"};
PermutationGenerator x = new PermutationGenerator(elements.length);
StringBuffer permutation;

while (x.hasMore())
{
permutation = new StringBuffer("%");
indices = x.getNext();
for (int i = 0; i < indices.length; i++) {
permutation.append(elements[indices[i]]).append("%");
}
System.out.println(permutation.toString());

}
}

}

先给你一个看看！