How to create a map of character and dynamic ArrayList in Java

Question

I want to create a hashmap defined as HashMap<Character,ArrayList<String>>. What I am trying to do is to read a set of strings which I am reading into a list. From this list I want to generate this hashmap. So if strings are something like this

Andy,Aman,Rocky,Ravi,Nick etc. Map should be like

A->[Andy,Aman]
R->[Rocky,Ravi]
N->Nick

What I have tried is something like this

ArrayList<String> xlist= new ArrayList<String>();
    ArrayList<String> list = new ArrayList<String>();
    HashMap<Character,ArrayList<String>> h = new HashMap<Character,ArrayList<String>>();
    Scanner s = new Scanner(System.in);
    for(String sin : list){
        Character x =sin.charAt(0);
        //System.out.println(sin.charAt(0));
        if(h.containsKey(x)){
            h.get(x).add(sin);
            //xlist.clear();
            //xlist = h.get(x);
            //xlist.add(sin);
            //h.put(x,xlist.clone());

        }
        else{
            xlist.clear();
            xlist.add(sin);
            h.put(x,xlist);

        }
    }

When I print the hashmap, I get this

{A=[Ravi, Rocky], R=[Ravi, Rocky], N=[Ravi, Rocky]}

I understand , that all the lists in values are being reflected from same copy of list, but I don't know , how to resolve it.

Kamil Chalczynski · Accepted Answer · 2016-06-23 00:20:55Z

3

If you don't want to rewrite the loop, you can just change your last line within the else block so you won't change content of map on every iteration:

h.put(x,new ArrayList<String>(Arrays.asList(sin)));

You can also do it using Java 8 Streams which makes it much shorter:

Map<Character, List<String>> map = list.stream().collect(Collectors.groupingBy(d -> d.charAt(0)));

If you want to preserve the order of keys you can use it with LinkedHashMap:

Map<Character, List<String>> map = list.stream().collect(Collectors.groupingBy(d->d.charAt(0), LinkedHashMap::new, Collectors.toList()));

edited Jun 23, 2016 at 0:20

answered Jun 22, 2016 at 23:36

Kamil Chalczynski

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1 Comment

Ravi Ranjan Singh Over a year ago

Thanks Kamil, that was so stupid of me, its looking so obvious now.

Andrew Rueckert · Accepted Answer · 2016-06-22 23:19:49Z

1

xlist is always the same list, even after you've put it in your hashmap. Whenever you hit a new letter, you clear every list in your hashmap, and add the current word to every list, because they're all the same list.

I'd rewrite your loop as:

for(String sin : list){
    Character x =sin.charAt(0);
    if(!h.containsKey(x)){
        h.put(x, new ArrayList<String>());
    }
    h.get(x).add(sin);
}

answered Jun 22, 2016 at 23:19

Andrew Rueckert

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2 Comments

shmosel Over a year ago

With Java 8, you can replace the whole loop body with h.computeIfAbsent(sin.charAt(0), c -> new ArrayList<>()).add(sin);

Ravi Ranjan Singh Over a year ago

@Andrew Ruekert: Thanks Kamil, that was so stupid of me, its looking so obvious now.Much cleaner loop now.

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