Question

我有一个字符串数组

A=["hello", "you"]

我有一个字符串，比如说

s="hello, hello you are so wonderful"

我需要计算A中来自s的字符串出现次数。在这种情况下，出现次数为3（2 "hello"，1 "you"）。

如何有效地做到这一点？（A可能包含大量单词，而s可能在实践中很长）

Answer 1

尝试：

Map<String, Integer> wordCount = new HashMap<>();
for(String a : dictionnary) {
    wordCount.put(a, 0);
}
for(String s : text.split("\\s+")) {
    Integer count = wordCount.get(s);
    if(count != null) {
        wordCount.put(s, count + 1);
    }
}

Answer 2

int count =0;
for(int i=0;i<A.length;i++)
{
  count = count + s.split(A[i],-1).length - 1;
}

工作Ideone：http://ideone.com/Z9K3JX

Answer 3

public void countMatches() {
    String[] A = {"hello", "you"};
    String s = "hello, hello you are so wonderful";
    String patternString = "(" + StringUtils.join(A, "|")   + ")";
    Pattern pattern = Pattern.compile(patternString);
    Matcher matcher = pattern.matcher(s);
    int count = 0;
    while (matcher.find()) {
        count++;
    }
    System.out.println(count);
}

请注意，StringUtils来自apache commons。如果你不想包含和使用额外的jar，你可以使用for循环来构造该字符串。

Answer 4

HashSet<String> searchWords = new HashSet<String>();

for(String a : dictionary) {
    searchWords.add(a);
}

int count = 0;

for(String s : input.split("[ ,]")) {
    if(searchWords.contains(s)) {
       count++;
    }
}

Answer 5

这是输出完全正常的方法：）

public static void main(String[] args) {
    String[] A={"hello", "you"};
    String s= "hello, hello you are so wonderful";
    int[] count = new int[A.length];
    for (int i = 0; i < A.length; i++) {
        count[i] = (s.length() - s.replaceAll(A[i], "").length())/A[i].length();
    }

    for (int i = 0; i < count.length; i++) {
        System.out.println(A[i] + ": " + count[i]);
    }
}

这条线做什么？

count[i] = (s.length() - s.replaceAll(A[i], "").length())/A[i].length();

此部分s.replaceAll(A[i], "")将所有“hello”更改为文本中的空“”字符串。

所以我考虑了所有内容的长度s.length()我从中减去了没有该词s.replaceAll(A[i], "").length()的同一个字符串的长度，并将其除以该字/A[i].length()的长度

此示例的示例输出：

hello: 2
you: 1

Answer 6

您可以使用String Tokenizer

做这样的事情：

A = ["hello", "you"];
s = "hello, hello you are so wonderful";
StringTokenizer st = new StringTokenizer(s);
    while (st.hasMoreElements()) {
        for (String i: A) {
            if(st.nextToken() == i){
                //You can keep going from here
            }
        }
}

Answer 7

这就是我提出的：

它不会创建任何新对象。它使用String.indexOf(String, int)，跟踪当前索引，并增加出现次数。

public class SearchWordCount  {
   public static final void main(String[] ignored)  {
      String[] searchWords = {"hello", "you"};
      String input = "hello, hello you are so wonderful";

      for(int i = 0; i < searchWords.length; i++)  {
         String searchWord = searchWords[i];

         System.out.print(searchWord + ": ");

         int foundCount = 0;
         int currIdx = 0;
         while(currIdx != -1)  {
            currIdx = input.indexOf(searchWord, currIdx);

            if(currIdx != -1)  {
               foundCount++;
               currIdx += searchWord.length();
            }  else  {
               currIdx = -1;
            }
         }

         System.out.println(foundCount);

      }
   }
}

输出：

hello: 2
you: 1

计算字符串中数组的单词

7 个答案: