号码分组逻辑

Question

我在不同的A，B和C组中有1 -10个。A-1、A-2、A-3、B-4、C-5、B-6、A-7、C-8、A-9、A、10

我想分成A、B、C三组

A
1-3,
7,
9-10

B
4,
6

C
5,
8

有没有人能帮我学逻辑？

Answer 1

这里有一些东西可以让你入门：

String[] items = new String[] {
  "A-1", "B-2", "A-5"
}

// This is the data structure that will receive the final data. The map key is the 
// group name (e.g. "A" for item "A-15") and the map value is a list of numbers that
// have been found for that group. TreeMap is chosen because the groups will be sorted
// alphabetically. If you don't need that, you could also use HashMap.
Map<String, List<Integer>> groups = new TreeMap<String, List<Integer>>();

for (String item : items) {
  // Split the item into the group and the number
  String group = item.substring(0, 1);
  String number = Integer.toString(item.substring(2));

  // See if this group is already registered in our Map
  List<Integer> groupData = groups.get(group);
  if (groupData==null) {
    groupData = new List<Integer>();
    groups.put(group, groupData);
  }

  // Add the number to the data
  groupData.add(number);
}

我在这里假设您的项目始终采用1字母破折号的形式。如果它比这复杂一点，你会想看看正则表达式(参见java类模式)。这是未测试的，我让您测试它并处理特殊情况。

此函数将为{ "A-1", "A-2", "A-3", "B-2", "A-5" }输出

A -> {1, 2, 3, 5}
B -> {2}

如果您想合并连续的数字，则需要处理生成的数字列表，但如果您仔细考虑一下，应该不会太困难。

Answer 2

Guava将帮助您创建所需的数据结构：

public static void main(final String[] args) {
    String input = "A-1,A-2,A-3,B-4,C-5,B-6,A-7,C-8,A-9,A-10";

    // create multimap
    Map<String, Collection<Integer>> map=Maps.newTreeMap();
    SortedSetMultimap<String, Integer> multimap = Multimaps.newSortedSetMultimap(
        map, new Supplier<SortedSet<Integer>>() {
          public SortedSet<Integer> get() {
              return new TreeSet<Integer>();
          }
    });

    //add data
    Splitter entrySplitter = Splitter.on(',');
    Splitter keyValueSplitter = Splitter.on('=');
    for (String entry : entrySplitter.split(input)) {
        Iterator<String> tokens = keyValueSplitter.split(entry).iterator();
        multimap.put(tokens.next(), Integer.valueOf(tokens.next()));
    }

    // read data
    for (Entry<String, Collection<Integer>> entry : map.entrySet()) {
        System.out.println(entry.getKey()+":");
        printMergedValues(entry.getValue());
    }

}

private static void printMergedValues(Collection<Integer> value) {
    // TODO implement this yourself
}

我留给你的唯一一件事就是加入小组

Answer 3

我会这样说：

     String[] input = {"A-1","A-2","A-3","B-4","C-5","B-6","A-7","C-8","A-9"};
     Map<String, Set<Integer>> result = new HashMap<String, Set<Integer>>();
     String[] inputSplit;
     String group;
     Integer groupNumber;
     for (String item : input)
     {
         inputSplit = item.split("-");
         group = inputSplit[0];
         groupNumber = Integer.valueOf( inputSplit[1] );
         if ( result.get(group) == null ) { result.put(group, new HashSet<Integer>()); }
         result.get(group).add(groupNumber);
     }

     for (Map.Entry entry : result.entrySet())
     {
         System.out.println( entry.getKey() + ":"  +  entry.getValue() );
     }