int count = (line.length() - line.replace("str", "").length())/"str".length();

使用lambda函数删除所有字符进行计数 计数是前长度和后长度之差

String s = "a.b.c.d";
int count = s.length() - deleteChars.apply( s, "." ).length();  // 3

在这里找到deleteChars

如果你需要计算多个字符的出现次数，你可以一次性完成: 如。对于b c和。

int count = s.length() - deleteChars.apply( s, "bc." ).length();  // 5

这是一个稍微不同风格的递归解决方案:

public static int countOccurrences(String haystack, char needle)
{
    return countOccurrences(haystack, needle, 0);
}

private static int countOccurrences(String haystack, char needle, int accumulator)
{
    if (haystack.length() == 0) return accumulator;
    return countOccurrences(haystack.substring(1), needle, haystack.charAt(0) == needle ? accumulator + 1 : accumulator);
}

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

也可以在Java 8中使用reduce来解决这个问题:

int res = "abdsd3$asda$asasdd$sadas".chars().reduce(0, (a, c) -> a + (c == '$' ? 1 : 0));
System.out.println(res);

输出:

3

如果你想数不。字符串'SELENIUM'中的相同字符，或者你想打印字符串'SELENIUM'中的唯一字符。

public class Count_Characters_In_String{

     public static void main(String []args){

        String s = "SELENIUM";
        System.out.println(s);
        int counter;

       String g = "";

        for( int i=0; i<s.length(); i++ ) { 

        if(g.indexOf(s.charAt(i)) == - 1){
           g=g+s.charAt(i); 
          }

       }
       System.out.println(g + " ");



        for( int i=0; i<g.length(); i++ ) {          
          System.out.print(",");

          System.out.print(s.charAt(i)+ " : ");
          counter=0; 
          for( int j=0; j<s.length(); j++ ) { 

        if( g.charAt(i) == s.charAt(j) ) {
           counter=counter+1;

           }      

          }
          System.out.print(counter); 
       }
     }
}

/******************** 输出 **********************/

硒

铒

S: 1, e: 2, l: 1, e: 1, n: 1, i: 1, u: 1