set

问题 I would like to get all the possible partitions (disjoint subsets of a set which union is the original set) of a multiset (some elements are equal and non-distinguishable from each other). Simpler case when one would like to yield the partitions of a simple set, in which there are no elements with multiplicity, in other words all elements are different. For this scenario I found this Ruby code on StackOwerflow which is very efficient, as not storing all the possible partitions, but yielding

问题 I have a List in which I need to add a prefix in all the elements of my list. Below is the way I am doing it by iterating the list and then adding it. Is there any other better way to do it? Any one-two liner that can do the same stuff? private static final List<DataType> DATA_TYPE = getTypes(); public static LinkedList<String> getData(TypeFlow flow) { LinkedList<String> paths = new LinkedList<String>(); for (DataType current : DATA_TYPE) { paths.add(flow.value() + current.value()); } return

问题 I have a List in which I need to add a prefix in all the elements of my list. Below is the way I am doing it by iterating the list and then adding it. Is there any other better way to do it? Any one-two liner that can do the same stuff? private static final List<DataType> DATA_TYPE = getTypes(); public static LinkedList<String> getData(TypeFlow flow) { LinkedList<String> paths = new LinkedList<String>(); for (DataType current : DATA_TYPE) { paths.add(flow.value() + current.value()); } return

问题 I have a List in which I need to add a prefix in all the elements of my list. Below is the way I am doing it by iterating the list and then adding it. Is there any other better way to do it? Any one-two liner that can do the same stuff? private static final List<DataType> DATA_TYPE = getTypes(); public static LinkedList<String> getData(TypeFlow flow) { LinkedList<String> paths = new LinkedList<String>(); for (DataType current : DATA_TYPE) { paths.add(flow.value() + current.value()); } return

问题 I have a List in which I need to add a prefix in all the elements of my list. Below is the way I am doing it by iterating the list and then adding it. Is there any other better way to do it? Any one-two liner that can do the same stuff? private static final List<DataType> DATA_TYPE = getTypes(); public static LinkedList<String> getData(TypeFlow flow) { LinkedList<String> paths = new LinkedList<String>(); for (DataType current : DATA_TYPE) { paths.add(flow.value() + current.value()); } return

问题 I have a List in which I need to add a prefix in all the elements of my list. Below is the way I am doing it by iterating the list and then adding it. Is there any other better way to do it? Any one-two liner that can do the same stuff? private static final List<DataType> DATA_TYPE = getTypes(); public static LinkedList<String> getData(TypeFlow flow) { LinkedList<String> paths = new LinkedList<String>(); for (DataType current : DATA_TYPE) { paths.add(flow.value() + current.value()); } return

问题 i have been trying to figure out an effective algorithm that returns a set such as it's intersection with the given sets isn't equal to {} . for example : let's say the given sets are {1,7,4},{2,8,5},{1,3},{2,6} the function must return the set {1,2} because it has an intersection point with all the given sets (the generated set needs to be as small as possible) 回答1: This is a brute force solution. Apparently, this is the well-known NP-complete problem Hitting Set. from itertools import

问题 i have been trying to figure out an effective algorithm that returns a set such as it's intersection with the given sets isn't equal to {} . for example : let's say the given sets are {1,7,4},{2,8,5},{1,3},{2,6} the function must return the set {1,2} because it has an intersection point with all the given sets (the generated set needs to be as small as possible) 回答1: This is a brute force solution. Apparently, this is the well-known NP-complete problem Hitting Set. from itertools import

问题 I need to find all the different intersections between two partitions of the same set. For example, if we have the following two partitions of the same set x = [[1, 2], [3, 4, 5], [6, 7, 8, 9, 10]] y = [[1, 3, 6, 7], [2, 4, 5, 8, 9, 10]] the required result is [[1], [2], [3], [4, 5], [6, 7], [8, 9, 10]]. In detail, we calculate the cartesian product between every subset of x and y, and for each of these products, we classify the elements in new subsets accordingly if they belong to the

问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 7 years ago . Improve this question I discovered this weird behavior of the removeAll method of AbstractSets when working with individual Comparators . Depending on the size of the compared collections a different comparator is used. It is actually documented in the API but I still cannot see the reason behind