给定未知数量的列表，每个列表的长度未知，我需要生成一个包含所有可能唯一组合的奇异列表。例如，给定以下列表：
X: [A, B, C]
Y: [W, X, Y, Z]
然后我应该能够生成 12 种组合：
[AW, AX, AY, AZ, BW, BX, BY, BZ, CW, CX, CY, CZ]
如果添加了 3 个元素的第三个列表，我将有 36 个组合，依此类推。
关于如何在 Java 中做到这一点的任何想法？
（伪代码也可以）
问问题
70524 次
11 回答
83
你需要递归：
假设您的所有列表都在 中lists，这是一个列表列表。让result成为您所需排列的列表。你可以像这样实现它：
void generatePermutations(List<List<Character>> lists, List<String> result, int depth, String current) {
if (depth == lists.size()) {
result.add(current);
return;
}
for (int i = 0; i < lists.get(depth).size(); i++) {
generatePermutations(lists, result, depth + 1, current + lists.get(depth).get(i));
}
}
最终的调用将是这样的：
generatePermutations(lists, result, 0, "");
于 2013-06-19T13:49:01.713 回答
32
这种操作称为笛卡尔积。Guava 为此提供了一个实用函数：Lists.cartesianProduct
于 2018-02-01T11:49:19.200 回答
21
这个话题就派上用场了。我已经完全用 Java 重写了以前的解决方案，并且更加用户友好。此外，我使用集合和泛型以获得更大的灵活性：
/**
* Combines several collections of elements and create permutations of all of them, taking one element from each
* collection, and keeping the same order in resultant lists as the one in original list of collections.
*
* <ul>Example
* <li>Input = { {a,b,c} , {1,2,3,4} }</li>
* <li>Output = { {a,1} , {a,2} , {a,3} , {a,4} , {b,1} , {b,2} , {b,3} , {b,4} , {c,1} , {c,2} , {c,3} , {c,4} }</li>
* </ul>
*
* @param collections Original list of collections which elements have to be combined.
* @return Resultant collection of lists with all permutations of original list.
*/
public static <T> Collection<List<T>> permutations(List<Collection<T>> collections) {
if (collections == null || collections.isEmpty()) {
return Collections.emptyList();
} else {
Collection<List<T>> res = Lists.newLinkedList();
permutationsImpl(collections, res, 0, new LinkedList<T>());
return res;
}
}
/** Recursive implementation for {@link #permutations(List, Collection)} */
private static <T> void permutationsImpl(List<Collection<T>> ori, Collection<List<T>> res, int d, List<T> current) {
// if depth equals number of original collections, final reached, add and return
if (d == ori.size()) {
res.add(current);
return;
}
// iterate from current collection and copy 'current' element N times, one for each element
Collection<T> currentCollection = ori.get(d);
for (T element : currentCollection) {
List<T> copy = Lists.newLinkedList(current);
copy.add(element);
permutationsImpl(ori, res, d + 1, copy);
}
}
我正在使用番石榴库来创建收藏。
于 2014-05-26T13:07:34.623 回答
9
添加基于迭代器的答案以适用于列表的通用列表List<List<T>>，扩展了 Ruslan Ostafiichuk 的答案的想法。我遵循的想法是：
* List 1: [1 2]
* List 2: [4 5]
* List 3: [6 7]
*
* Take each element from list 1 and put each element
* in a separate list.
* combinations -> [ [1] [2] ]
*
* Set up something called newCombinations that will contains a list
* of list of integers
* Consider [1], then [2]
*
* Now, take the next list [4 5] and iterate over integers
* [1]
* add 4 -> [1 4]
* add to newCombinations -> [ [1 4] ]
* add 5 -> [1 5]
* add to newCombinations -> [ [1 4] [1 5] ]
*
* [2]
* add 4 -> [2 4]
* add to newCombinations -> [ [1 4] [1 5] [2 4] ]
* add 5 -> [2 5]
* add to newCombinations -> [ [1 4] [1 5] [2 4] [2 5] ]
*
* point combinations to newCombinations
* combinations now looks like -> [ [1 4] [1 5] [2 4] [2 5] ]
* Now, take the next list [6 7] and iterate over integers
* ....
* 6 will go into each of the lists
* [ [1 4 6] [1 5 6] [2 4 6] [2 5 6] ]
* 7 will go into each of the lists
* [ [1 4 6] [1 5 6] [2 4 6] [2 5 6] [1 4 7] [1 5 7] [2 4 7] [2 5 7]]
现在是代码。我用了一个Set简单的来摆脱任何重复。可以替换为List. 一切都应该无缝地工作。:)
public static <T> Set<List<T>> getCombinations(List<List<T>> lists) {
Set<List<T>> combinations = new HashSet<List<T>>();
Set<List<T>> newCombinations;
int index = 0;
// extract each of the integers in the first list
// and add each to ints as a new list
for (T i : lists.get(0)) {
List<T> newList = new ArrayList<T>();
newList.add(i);
combinations.add(newList);
}
index++;
while (index < lists.size()) {
List<T> nextList = lists.get(index);
newCombinations = new HashSet<List<T>>();
for (List<T> first : combinations) {
for (T second : nextList) {
List<T> newList = new ArrayList<T>();
newList.addAll(first);
newList.add(second);
newCombinations.add(newList);
}
}
combinations = newCombinations;
index++;
}
return combinations;
}
一个小测试块..
public static void main(String[] args) {
List<Integer> l1 = Arrays.asList(1, 2, 3);
List<Integer> l2 = Arrays.asList(4, 5);
List<Integer> l3 = Arrays.asList(6, 7);
List<List<Integer>> lists = new ArrayList<List<Integer>>();
lists.add(l1);
lists.add(l2);
lists.add(l3);
Set<List<Integer>> combs = getCombinations(lists);
for (List<Integer> list : combs) {
System.out.println(list.toString());
}
}
于 2016-02-26T13:06:37.423 回答
7
没有递归唯一组合：
String sArray[] = new String[]{"A", "A", "B", "C"};
//convert array to list
List<String> list1 = Arrays.asList(sArray);
List<String> list2 = Arrays.asList(sArray);
List<String> list3 = Arrays.asList(sArray);
LinkedList<List<String>> lists = new LinkedList<List<String>>();
lists.add(list1);
lists.add(list2);
lists.add(list3);
Set<String> combinations = new TreeSet<String>();
Set<String> newCombinations;
for (String s : lists.removeFirst())
combinations.add(s);
while (!lists.isEmpty()) {
List<String> next = lists.removeFirst();
newCombinations = new TreeSet<String>();
for (String s1 : combinations)
for (String s2 : next)
newCombinations.add(s1 + s2);
combinations = newCombinations;
}
for (String s : combinations)
System.out.print(s + " ");
于 2013-06-19T14:56:09.743 回答
5
您需要实现的操作称为笛卡尔积。有关更多详细信息，请参阅https://en.wikipedia.org/wiki/Cartesian_product
我建议使用我的开源库，它可以完全满足您的需求： https ://github.com/SurpSG/Kombi
有如何使用它的例子： https ://github.com/SurpSG/Kombi#usage-for-lists-1
注意：该库是为高性能目的而设计的。您可以在此处观察基准结果
该库为您提供了非常好的吞吐量和恒定的内存使用
于 2018-04-18T18:46:14.780 回答
3
使用此处其他一些答案提供的嵌套循环解决方案来组合两个列表。
当您有两个以上的列表时，
- 将前两个组合成一个新列表。
- 将结果列表与下一个输入列表合并。
- 重复。
于 2013-06-19T13:48:36.483 回答
3
- 没有递归
- 已订购
- 可以通过其索引获得特定组合（无需构建所有其他排列）
- 支持并行迭代
到底类和main()方法：
public class TwoDimensionalCounter<T> {
private final List<List<T>> elements;
private final int size;
public TwoDimensionalCounter(List<List<T>> elements) {
//Need to reverse the collection if you need the original order in the answers
this.elements = Collections.unmodifiableList(elements);
int size = 1;
for(List<T> next: elements)
size *= next.size();
this.size = size;
}
public List<T> get(int index) {
List<T> result = new ArrayList<>();
for(int i = elements.size() - 1; i >= 0; i--) {
List<T> counter = elements.get(i);
int counterSize = counter.size();
result.add(counter.get(index % counterSize));
index /= counterSize;
}
return result;
}
public int size() {
return size;
}
public static void main(String[] args) {
TwoDimensionalCounter<Integer> counter = new TwoDimensionalCounter<>(
Arrays.asList(
Arrays.asList(1, 2, 3),
Arrays.asList(1, 2),
Arrays.asList(1, 2, 3)
));
for(int i = 0; i < counter.size(); i++)
System.out.println(counter.get(i));
}
}
PS：事实证明，番石榴的笛卡尔积使用相同的算法。但是他们还为 List 创建了特殊的子类，以使其效率提高几倍。
于 2017-07-17T11:49:03.400 回答
3
像往常一样迟到，但这里有一个很好解释的使用数组的例子。它可以很容易地更改为列表。我需要按字典顺序为我的用例提供多个数组的所有唯一组合。
我发布它是因为这里的答案都没有给出明确的算法，而且我无法忍受递归。毕竟我们不在stackoverflow上吗？
String[][] combinations = new String[][] {
new String[] { "0", "1" },
new String[] { "0", "1" },
new String[] { "0", "1" },
new String[] { "0", "1" } };
int[] indices = new int[combinations.length];
int currentIndex = indices.length - 1;
outerProcess: while (true) {
for (int i = 0; i < combinations.length; i++) {
System.out.print(combinations[i][indices[i]]);
}
System.out.println();
while (true) {
// Increase current index
indices[currentIndex]++;
// If index too big, set itself and everything right of it to 0 and move left
if (indices[currentIndex] >= combinations[currentIndex].length) {
for (int j = currentIndex; j < indices.length; j++) {
indices[j] = 0;
}
currentIndex--;
} else {
// If index is allowed, move as far right as possible and process next
// combination
while (currentIndex < indices.length - 1) {
currentIndex++;
}
break;
}
// If we cannot move left anymore, we're finished
if (currentIndex == -1) {
break outerProcess;
}
}
}
输出;
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
于 2018-02-26T12:47:53.947 回答
1
使用Java 8 Stream map和reduce方法生成组合。
public static <T> List<List<T>> combinations(List<List<T>> lists) {
// incorrect incoming data
if (lists == null) return Collections.emptyList();
return lists.stream()
// non-null and non-empty lists
.filter(list -> list != null && list.size() > 0)
// represent each list element as a singleton list
.map(list -> list.stream().map(Collections::singletonList)
// Stream<List<List<T>>>
.collect(Collectors.toList()))
// summation of pairs of inner lists
.reduce((list1, list2) -> list1.stream()
// combinations of inner lists
.flatMap(inner1 -> list2.stream()
// merge two inner lists into one
.map(inner2 -> Stream.of(inner1, inner2)
.flatMap(List::stream)
.collect(Collectors.toList())))
// list of combinations
.collect(Collectors.toList()))
// otherwise an empty list
.orElse(Collections.emptyList());
}
public static void main(String[] args) {
List<String> list1 = Arrays.asList("A", "B", "C");
List<String> list2 = Arrays.asList("W", "X", "Y", "Z");
List<String> list3 = Arrays.asList("L", "M", "K");
List<List<String>> lists = Arrays.asList(list1, list2, list3);
List<List<String>> combinations = combinations(lists);
// column-wise output
int rows = 6;
IntStream.range(0, rows).forEach(i -> System.out.println(
IntStream.range(0, combinations.size())
.filter(j -> j % rows == i)
.mapToObj(j -> combinations.get(j).toString())
.collect(Collectors.joining(" "))));
}
逐列输出：
[A, W, L] [A, Y, L] [B, W, L] [B, Y, L] [C, W, L] [C, Y, L]
[A, W, M] [A, Y, M] [B, W, M] [B, Y, M] [C, W, M] [C, Y, M]
[A, W, K] [A, Y, K] [B, W, K] [B, Y, K] [C, W, K] [C, Y, K]
[A, X, L] [A, Z, L] [B, X, L] [B, Z, L] [C, X, L] [C, Z, L]
[A, X, M] [A, Z, M] [B, X, M] [B, Z, M] [C, X, M] [C, Z, M]
[A, X, K] [A, Z, K] [B, X, K] [B, Z, K] [C, X, K] [C, Z, K]
另请参阅：任意数量集合的笛卡尔积
于 2021-04-21T11:44:25.730 回答
-5
这是使用位掩码的示例。没有递归和多个列表
static List<Integer> allComboMatch(List<Integer> numbers, int target) {
int sz = (int)Math.pow(2, numbers.size());
for (int i = 1; i < sz; i++) {
int sum = 0;
ArrayList<Integer> result = new ArrayList<Integer>();
for (int j = 0; j < numbers.size(); j++) {
int x = (i >> j) & 1;
if (x == 1) {
sum += numbers.get(j);
result.add(j);
}
}
if (sum == target) {
return result;
}
}
return null;
}
于 2017-11-17T20:56:56.327 回答