我知道之前也曾问过类似的问题,但我发现答案令人困惑。我正在尝试制作一个程序,它将找到一个没有重复且只有最大大小的数组列表的每个组合。如果列表有4个项目,则应仅打印出所有4个项目均存在的组合。这是我到目前为止的内容:
public main(){
UI.initialise();
UI.addButton("Test", this::testCreate);
UI.addButton("Quit", UI::quit);
}
public void createCombinations(ArrayList<String> list, String s, int depth) {
if (depth == 0) {
return;
}
depth --;
for (int i = 0; i < list.size(); i++) {
if (this.constrain(s + "_" + list.get(i), list.size())) {
UI.println(s + "_" + list.get(i));
}
createCombinations(list, s + "_" + list.get(i), depth);
}
}
public void testCreate() {
ArrayList<String> n = new ArrayList<String>();
n.add("A"); n.add("B"); n.add("C"); n.add("D");
this.createCombinations(n , "", n.size());
}
public boolean constrain(String s, int size) {
// Constrain to only the maximum length
if ((s.length() != size*2)) {
return false;
}
// Constrain to only combinations without repeats
Scanner scan = new Scanner(s).useDelimiter("_");
ArrayList<String> usedTokens = new ArrayList<String>();
String token;
while (scan.hasNext()) {
token = scan.next();
if (usedTokens.contains(token)) {
return false;
} else {
usedTokens.add(token);
}
}
// If we fully iterate over the loop then there are no repitions
return true;
}
public static void main(String[] args){
main obj = new main();
}
这会打印出正确的以下内容:
_A_B_C_D
_A_B_D_C
_A_C_B_D
_A_C_D_B
_A_D_B_C
_A_D_C_B
_B_A_C_D
_B_A_D_C
_B_C_A_D
_B_C_D_A
_B_D_A_C
_B_D_C_A
_C_A_B_D
_C_A_D_B
_C_B_A_D
_C_B_D_A
_C_D_A_B
_C_D_B_A
_D_A_B_C
_D_A_C_B
_D_B_A_C
_D_B_C_A
_D_C_A_B
_D_C_B_A
这对于较小的列表有效,但对于较大的列表效率很低。我知道我所做的事情是完全错误的,但是我想学习正确的方法。任何帮助都非常感谢。提前致谢。
P.S。这不是家庭作业,只是出于兴趣,尽管我是一名新的CS学生(如果不是很明显)。
答案 0 :(得分:0)
在Java中实现Heap's algorithm:
import java.util.Arrays;
public class Main {
public static void swap(final Object[] array, final int index1, final int index2) {
final Object tmp = array[index1];
array[index1] = array[index2];
array[index2] = tmp;
}
public static void printPermutations_HeapsAlgorithm(final int n, final Object[] array) {
final int[] c = new int[n];
for (int i = 0; i < c.length; ++i)
c[i] = 0;
System.out.println(Arrays.toString(array)); //Consume first permutation.
int i=0;
while (i < n) {
if (c[i] < i) {
if ((i & 1) == 0)
swap(array, 0, i);
else
swap(array, c[i], i);
System.out.println(Arrays.toString(array)); //Consume permutation.
++c[i];
i=0;
}
else
c[i++] = 0;
}
}
public static void main(final String[] args) {
printPermutations_HeapsAlgorithm(4, new Character[]{'A', 'B', 'C', 'D'});
}
}
可能是this的副本。