如何使用Java流使列表元素的所有变化?

Question

我有一个列表，例如:

["a", "b", "c"]

，我想创建一个列表的列表，它将是这个列表元素的变体。结果应该是这样的:

[["a"], 
["b"], 
["c"], 
["a", "b"], 
["a", "c"], 
["b", "a"], 
["b", "c"], 
["c", "a"], 
["c", "b"], 
["a", "b", "c"], 
["a", "c", "b"], 
["b", "a", "c"], 
["b", "c", "a"], 
["c", "a", "b"], 
["c", "b", "a"]]

最好(但不是必须)的解决方案是使用java流。

我正在尝试这样做:

List<List<String>> lists = IntStream.rangeClosed(1, list.size()) //size of sublist
.flatMap(i -> list.subList(0, i).stream()).collect(Collectors.toList())

但是不能用

Answer 1

你需要做的是利用k-置换。您需要对传递的数组的每个长度执行k置换，以获得所有可能的变化。

:

private static List<List<String>> getAllVariations(List<String> elements){
return IntStream.rangeClosed(1, elements.size()).boxed()
.flatMap(i-> kPermutate(elements,0,i).stream())
.collect(Collectors.toList());
}
private static List<List<String>> kPermutate(List<String> allElements, int i, int numberOfElements)
{
if(i == numberOfElements)
{
//New ArrayList because we want a clone of the sublist
return Arrays.asList(new ArrayList<>(allElements.subList(0, numberOfElements)));
}
List<List<String>> tempResult = new ArrayList<>();
for(int j=i; j<allElements.size(); j++)
{
Collections.swap(allElements, i, j);
tempResult.addAll(kPermutate(allElements, i+1, numberOfElements));
Collections.swap(allElements, i, j);
}
return tempResult;
}

测试:

public static void main(String[] args) {
List<String> elements = Arrays.asList("a", "b", "c");
List<List<String>> result = getAllVariations(elements);
System.out.println(result);
}

输出:

[[a], [b], [c], [a, b], [a, c], [b, a], [b, c], [c, b], [c, a], [a, b, c], [a, c, b], [b, a, c], [b, c, a], [c, b, a], [c, a, b]]

如果你不想在getallvariables方法中使用流，你也可以像这样将它实现为常规的for循环

private static List<List<String>> getAllVariations(List<String> elements){
List<List<String>> result = new ArrayList<>();
//Start from 1 because we don't want to include [] in the list.
//i<=elements.size(): we use <= because we want the max number of permutations. this is not a 0 based index for that
for(int i=1;i<=elements.size();i++){
result.addAll(kPermutate(elements,0,i));
}
return result;
}