如何获取元素在ArrayList中出现的次数

Question

    for(int i=0; i<arr.size(); ++i){
        oc = Collections.frequency(arr, arr.get(i));
        System.out.print(oc + " "+ arr.get(i) +" ");
        arr.remove(i);
    }

The main idea is to output how many times every element in sequence occurs. For example

1 1 2 3 3 3 10 10

here the output is

2 1 1 2 3 3 2 10

It is like two ones, one element of two, 3 elements of 3, and 2 elements of 10.

This is buggy, for example it is not working for this case

1 1 1 2 2 1 1 1 

What is wrong? Any other algorithms?

Answer 1

The problem is that inside the for loop you remove an element ( arr.remove(i) ) so the remaining elements get shifted and when i gets incremented, you skip one element. Removing an element also changes its frequency, so don't do that.

Do something like this:

List<String> arr = Arrays.asList("a", "a", "b", "a", "a");

for (String s : arr)
    System.out.println("element: " + s
        + ", count: " + Collections.frequency(arr, s));

If an element is in the list multiple times, this will print it multiple times. Use a HashSet to remember if an element was already printed, and do not print it again:

List<String> arr = Arrays.asList("a", "a", "b", "a", "a");

Set<String> printed = new HashSet<>();
for (String s : arr) {
    if (printed.add(s)) // Set.add() also tells if the element was in the Set!
        System.out.println("element: " + s
            + ", count: " + Collections.frequency(arr, s));
}

Output:

element: a, count: 4
element: b, count: 1

Alternative

Alternatively you can add all elements of the original list to a Set (which will ensure every element is contained only once), and iterate over this set, but count in the original array:

List<String> arr = Arrays.asList("a", "a", "b", "a", "a");

for (String s : new HashSet<>(arr))
    System.out.println("element: " + s
        + ", count: " + Collections.frequency(arr, s));

Output: same. But note that this might result in different order of the output as Set s in Java are not ordered.

Answer 2

I would work with a HashMap, whereas the meaning is element -> count! pseudo:

HashMap<Integer, Integer> counts = new HashMap<Integer, Integer>();
 for(int i=0; i<arr.size(); ++i){
    Integer x = counts.get(arr.get(i));
    if (x==null) counts.put(arr.get(i), 1);
    else counts.put(arr.get(i), x+1);
}

after this, your hashmap holds all elements and their count

Answer 3

try this

    List a = Arrays.asList(1, 2, 1, 3, 1);
    Collections.sort(a);
    Object o = a.get(0);
    int n = 1;
    for (int i = 1; i < a.size(); i++) {
        Object t = a.get(i);
        if (o.equals(t)) {
            n++;
        } else {
            System.out.println(o + " - " + n);
            n = 1;
            o = t;
        }
    }
    System.out.println(o + " - " + n);

output

1 - 3
2 - 1
3 - 1

Answer 4

快速，聪明的方法是：1）使用Collections.sort对arrayList进行排序2）使用indexOf（）获取第一个索引，并使用lastIndexOf（）方法获取最后一个索引3）2个索引的差将为您提供出现次数给定对象在ArrayList中。