如果我理解正确，你需要生成所有的 64！ 64字节数组的排列，即：
64！ = 126886932185884164103433389335161480802865516174545192198801894375214704230400000000000000排列！
如果每个排列和比较都需要一毫秒（最坏的时间场景）来计算，您需要：
4023558225072430368576654912961741527234446859923426946943236123009091331506849.3150684931506849315 年在一台机器上计算它们！ （如果每个排列都需要 100 毫秒，那么这个怪物的 100 分之一）。
因此，您应该通过应用一些启发式方法来减少问题的搜索空间，而不是天真地列出所有可能的解决方案。
在您将搜索空间缩小为更易于处理的数字后，例如：14！ （“一毫秒”情况下的计算时间为 2 年），您可以使用 Factoradics（实现 here）将计算拆分到多台机器上，计算每台机器的开始和结束排列，然后使用每个节点中的以下代码（Knuth's L-algorithm 的实现）在每台机器中搜索解决方案：
public class Perm {
private static byte[] sequenceToMatch;
private static byte[] startSequence;
private static byte[] endingSequence;
private static final int SEQUENCE_LENGTH = 64;
public static void main(String... args) {
final int N = 3;
startSequence = readSequence(args[0]);
endingSequence = readSequence(args[1]);
sequenceToMatch = readSequence(args[2]);
permutations();
}
private static boolean sequencesMatch(byte[] s1, byte[] s2) {
for (int i = 0; i < SEQUENCE_LENGTH; i++) {
if (s1[i] != s2[i]) {
return false;
}
}
return true;
}
private static byte[] readSequence(String argument) {
String[] sBytes = argument.split(",");
byte[] bytes = new byte[SEQUENCE_LENGTH];
int i = 0;
for (String sByte : sBytes) {
bytes[i++] = Byte.parseByte(sByte, 10);
}
return bytes;
}
private static void swap(byte[] elements, int i, int j) {
byte temp = elements[i];
elements[i] = elements[j];
elements[j] = temp;
}
/**
* Reverses the elements of an array (in place) from the start index to the end index
*/
private static void reverse(byte[] array, int startIndex, int endIndex) {
int size = endIndex + 1 - startIndex;
int limit = startIndex + size / 2;
for (int i = startIndex; i < limit; i++) {
// swap(array, i, startIndex + (size - 1 - (i - startIndex)));
swap(array, i, 2 * startIndex + size - 1 - i);
}
}
/**
* Implements the Knuth's L-Algorithm permutation algorithm
* modifying the collection in place
*/
private static void permutations() {
byte[] sequence = startSequence;
if (sequencesMatch(sequence, sequenceToMatch)) {
System.out.println("Solution found!");
return;
}
// For every possible permutation
while (!sequencesMatch(sequence, endingSequence)) {
// Iterate the array from right to left in search
// of the first couple of elements that are in ascending order
for (int i = SEQUENCE_LENGTH - 1; i >= 1; i--) {
// If the elements i and i - 1 are in ascending order
if (sequence[i - 1] < sequence[i]) {
// Then the index "i - 1" becomes our pivot index
int pivotIndex = i - 1;
// Scan the elements at the right of the pivot (again, from right to left)
// in search of the first element that is bigger
// than the pivot and, if found, swap it
for (int j = SEQUENCE_LENGTH - 1; j > pivotIndex; j--) {
if (sequence[j] > sequence[pivotIndex]) {
swap(sequence, j, pivotIndex);
break;
}
}
// Now reverse the elements from the right of the pivot index
// (this nice touch to the algorithm avoids the recursion)
reverse(sequence, pivotIndex + 1, SEQUENCE_LENGTH - 1);
break;
}
}
if (sequencesMatch(sequence, sequenceToMatch)) {
System.out.println("Solution found!");
return;
}
}
}
}