使用Java 8和HashMap，没有任何库来计数所有不同的字符:

private static void countChars(String string) {
    HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
    string.chars().forEach(letter -> hm.put(letter, (hm.containsKey(letter) ? hm.get(letter) : 0) + 1));
    hm.forEach((c, i) -> System.out.println(((char)c.intValue()) + ":" + i));
}

好的，受Yonatan解决方案的启发，这里有一个纯递归的库方法-唯一使用的库方法是length()和charAt()，它们都不做任何循环:

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

private static int countOccurrences(String haystack, char needle, int index)
{
    if (index >= haystack.length())
    {
        return 0;
    }

    int contribution = haystack.charAt(index) == needle ? 1 : 0;
    return contribution + countOccurrences(haystack, needle, index+1);
}

递归是否算作循环取决于您使用的确切定义，但这可能是您能得到的最接近的定义。

我不知道现在大多数jvm是否都做尾递归…当然，如果不是这样，对于适当长的字符串就会出现同名堆栈溢出。

我试图用一个开关语句来解决你的问题，但我仍然需要一个for循环来解析字符串。如果我可以改进代码，请随意评论

public class CharacterCount {
public static void main(String args[])
{
    String message="hello how are you";
    char[] array=message.toCharArray();
    int a=0;
    int b=0;
    int c=0;
    int d=0;
    int e=0;
    int f=0;
    int g=0;
    int h=0;
    int i=0;
    int space=0;
    int j=0;
    int k=0;
    int l=0;
    int m=0;
    int n=0;
    int o=0;
    int p=0;
    int q=0;
    int r=0;
    int s=0;
    int t=0;
    int u=0;
    int v=0;
    int w=0;
    int x=0;
    int y=0;
    int z=0;


    for(char element:array)
    {
        switch(element)
        {
        case 'a':
        a++;
        break;
        case 'b':
        b++;
        break;
        case 'c':c++;
        break;

        case 'd':d++;
        break;
        case 'e':e++;
        break;
        case 'f':f++;
        break;

        case 'g':g++;
        break;
        case 'h':
        h++;
        break;
        case 'i':i++;
        break;
        case 'j':j++;
        break;
        case 'k':k++;
        break;
        case 'l':l++;
        break;
        case 'm':m++;
        break;
        case 'n':m++;
        break;
        case 'o':o++;
        break;
        case 'p':p++;
        break;
        case 'q':q++;
        break;
        case 'r':r++;
        break;
        case 's':s++;
        break;
        case 't':t++;
        break;
        case 'u':u++;
        break;
        case 'v':v++;
        break;
        case 'w':w++;
        break;
        case 'x':x++;
        break;
        case 'y':y++;
        break;
        case 'z':z++;
        break;
        case ' ':space++;
        break;
        default :break;
        }
    }
    System.out.println("A "+a+" B "+ b +" C "+c+" D "+d+" E "+e+" F "+f+" G "+g+" H "+h);
    System.out.println("I "+i+" J "+j+" K "+k+" L "+l+" M "+m+" N "+n+" O "+o+" P "+p);
    System.out.println("Q "+q+" R "+r+" S "+s+" T "+t+" U "+u+" V "+v+" W "+w+" X "+x+" Y "+y+" Z "+z);
    System.out.println("SPACE "+space);
}

}

public static void getCharacter(String str){

        int count[]= new int[256];

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


            count[str.charAt(i)]++;

        }
        System.out.println("The ascii values are:"+ Arrays.toString(count));

        //Now display wht character is repeated how many times

        for (int i = 0; i < count.length; i++) {
            if (count[i] > 0)
               System.out.println("Number of " + (char) i + ": " + count[i]);
        }


    }
}

public class OccurencesInString { public static void main(String[] args) { String str = "NARENDRA AMILINENI"; HashMap occur = new HashMap(); int count =0; String key = null; for(int i=0;i<str.length()-1;i++){ key = String.valueOf(str.charAt(i)); if(occur.containsKey(key)){ count = (Integer)occur.get(key); occur.put(key,++count); }else{ occur.put(key,1); } } System.out.println(occur); } }

那么下面的递归算法呢?这也是线性时间。

import java.lang.*;
import java.util.*;

class longestSubstr{

public static void main(String[] args){
   String s="ABDEFGABEF";


   int ans=calc(s);

   System.out.println("Max nonrepeating seq= "+ans);

}

public static int calc(String s)
{//s.s
      int n=s.length();
      int max=1;
      if(n==1)
          return 1;
      if(n==2)
      {
          if(s.charAt(0)==s.charAt(1)) return 1;
          else return 2;


      }
      String s1=s;
    String a=s.charAt(n-1)+"";
          s1=s1.replace(a,"");
         // System.out.println(s+" "+(n-2)+" "+s.substring(0,n-1));
         max=Math.max(calc(s.substring(0,n-1)),(calc(s1)+1));


return max;
}


}


</i>