得到答案最简单的方法如下:

public static void main(String[] args) {
    String string = "a.b.c.d";
    String []splitArray = string.split("\\.",-1);
    System.out.println("No of . chars is : " + (splitArray.length-1));
}

 public static int countSubstring(String subStr, String str) {

    int count = 0;
    for (int i = 0; i < str.length(); i++) {
        if (str.substring(i).startsWith(subStr)) {
            count++;
        }
    }
    return count;
}

灵感来自Jon Skeet，一个非循环版本，不会吹你的堆栈。如果你想使用fork-join框架，这也是一个有用的起点。

public static int countOccurrences(CharSequeunce haystack, char needle) {
    return countOccurrences(haystack, needle, 0, haystack.length);
}

// Alternatively String.substring/subsequence use to be relatively efficient
//   on most Java library implementations, but isn't any more [2013].
private static int countOccurrences(
    CharSequence haystack, char needle, int start, int end
) {
    if (start == end) {
        return 0;
    } else if (start+1 == end) {
        return haystack.charAt(start) == needle ? 1 : 0;
    } else {
        int mid = (end+start)>>>1; // Watch for integer overflow...
        return
            countOccurrences(haystack, needle, start, mid) +
            countOccurrences(haystack, needle, mid, end);
    }
}

(免责声明:未经测试，未经编译，不合理。)

也许最好的(单线程，不支持代理对)编写方法是:

public static int countOccurrences(String haystack, char needle) {
    int count = 0;
    for (char c : haystack.toCharArray()) {
        if (c == needle) {
           ++count;
        }
    }
    return count;
}

我对此的“惯用语”是:

int count = StringUtils.countMatches("a.b.c.d", ".");

既然已经是通用语言了，为什么还要自己写呢?

Spring Framework的线性程序是:

int occurance = StringUtils.countOccurrencesOf("a.b.c.d", ".");

在代码的某个地方，某些东西必须循环。解决这个问题的唯一方法是完全展开循环:

int numDots = 0;
if (s.charAt(0) == '.') {
    numDots++;
}

if (s.charAt(1) == '.') {
    numDots++;
}


if (s.charAt(2) == '.') {
    numDots++;
}

.．.等等，但你是在源代码编辑器中手动执行循环的人——而不是运行它的计算机。请看伪代码:

create a project
position = 0
while (not end of string) {
    write check for character at position "position" (see above)
}
write code to output variable "numDots"
compile program
hand in homework
do not think of the loop that your "if"s may have been optimized and compiled to

lambda一行代码 不需要外部库。 用每个字符的计数创建一个映射:

Map<Character,Long> counts = "a.b.c.d".codePoints().boxed().collect(
    groupingBy( t -> (char)(int)t, counting() ) );

获取:{a=1, b=1, c=1, d=1， .=3} 某一人物的数目。”。’已经结束了: 计数。('。”)

(出于病态的好奇心，我也写了一个lambda解，想知道我的解有多慢，最好是来自有10行解的人。)