计算字符的所有出现（字符串末尾除外）

Question

For example

String text = "sentence";     // (number of e's = 2)

There are three e's in that string, but the result should be 2 because the third one is at the very end. This is what I have so far:

public static void main(String[] args) {
    int count =0;
    String text    = "sentence";
    Pattern pat = Pattern.compile("[e]+");
    Matcher m = pat.matcher(text);
    while (m.find()) {
        count++;
    }
    System.out.println(count);
}

Answer 1

Replace + which exists after e with negative lookahead. e+ matches one or more e 's , so regex engine should consider eee as single match. And a negative lookahead after e , ie e(?!$) helps to find all the e 's but not the one which exists at the end of a line.

int count = 0;
String text = "sentence";
Pattern pat = Pattern.compile("e(?!$)");
Matcher m = pat.matcher(text);
while (m.find()) {
        count++;
}
System.out.println(count);

Answer 2

Matcher methods can tell you the start and end index of the match. If end (next character after) matches the length of the string then it's the last character. Eg

int count =0;
String text = "sentence";
Pattern pat = Pattern.compile("e");
Matcher m = pat.matcher(text);
while (m.find() && m.end() != text.length()) {
    count++;
}
System.out.println(count);

If you would like to exclude from counting last letter of a word instead of last word of the sentence, you can check whether the end character is alpha :

int count =0;
String text = "sentence";
Pattern pat = Pattern.compile("e");
Matcher m = pat.matcher(text);
while (m.find() &&
       m.end() != text.length() &&
       Character.isAlphabetic(text.charAt(m.end()))) {
    count++;
}
System.out.println(count);