因此,我正在构建一个霍夫曼树,需要将一个String用作输入,然后创建2个数组,其中包含每个字母以及该字母在原始字符串中的出现次数,如下所示:
String s = "mississippi"
应导致:
char[] charArr = {'m','i', 's', 'p'};
int[] count = {1,4,4,2};
关于此有很多问题,还有很多有关如何解决此问题的示例,尤其是在stackoverflow上,但是我设法做到的唯一一个是:
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;
}
}
但这感觉很麻烦,而且还会打乱数组中字母的顺序。如果我使用这个,我得到的数组如下:
char[] charArr = {'p','s', 'i', 'm'};
有什么更好的方法可以做我想要的吗?
最佳答案
我会这样
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
关于java - 计算出现次数并从字符串中删除重复项,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/41189214/