java - 计算字符串中的出现次数并删除重复项

Question

所以我正在构建一棵哈夫曼树，我需要将一个字符串作为输入，然后创建两个数组，其中包含每个字母以及该字母在原始字符串中出现的次数，如下所示:

String s = "mississippi"

应该导致:

char[] charArr = {'m','i', 's', 'p'};
int[] count = {1,4,4,2};

关于这个问题有很多问题，也有很多关于如何解决这个问题的例子，特别是在 stackoverflow 上，但我唯一能解决的问题是:

private void findOccurences(String s) {
        List<Character> original = new ArrayList<Character>(s.length());
        List<Character> duplicateRemoved;

        for (int i = 0; i < s.length(); i++) {
            original.add(s.charAt(i));
        }
        duplicateRemoved = new ArrayList<Character>(original);

        // Remove duplicates from second list.
        Set<Character> hs = new HashSet<Character>();
        hs.addAll(duplicateRemoved);
        duplicateRemoved.clear();
        duplicateRemoved.addAll(hs);

        charFreqs = new int[duplicateRemoved.size()];
        charArr = new char[duplicateRemoved.size()];

        for (int i = 0; i < charArr.length; i++) {
            char c = duplicateRemoved.get(i);
            int count = Collections.frequency(original, c);
            charArr[i] = c;
            charFreqs[i] = count;
        }
    }

但是感觉很麻烦，而且还打乱了数组中字母的顺序。如果我使用它，我的结果数组如下:

char[] charArr = {'p','s', 'i', 'm'};

有没有更好的方法来做我想做的事？

Answer 1

我会这样做

String s = "mississippi";
List<String> original = Arrays.stream(s.split(""))
                     .collect(Collectors.toList());
List<String> duplicateRemoved = Arrays.stream(s.split(""))
                      .distinct()
                      .collect(Collectors.toList());
ArrayList<Integer> Occurrences = new ArrayList<>();
int counter = 1;

for (String aList : duplicateRemoved) {
    counter = (int) original.stream().filter(s1 -> s1.equals(aList)).count();
            Occurrences.add(counter);
        }
System.out.println(duplicateRemoved);
System.out.println(Occurrences);

和输出