我有字符串

a.b.c.d

我想数一下'的出现次数。,最好是一句单句俏皮话。

(之前我把这个约束表述为“不使用循环”,以防你想知道为什么每个人都试图在不使用循环的情况下回答)。


当前回答

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);
}

其他回答

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", ".");

一个简短的例子是

String text = "a.b.c.d";
int count = text.split("\\.",-1).length-1;

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);
}

完整的示例:

public class CharacterCounter
{

  public static int countOccurrences(String find, String string)
  {
    int count = 0;
    int indexOf = 0;

    while (indexOf > -1)
    {
      indexOf = string.indexOf(find, indexOf + 1);
      if (indexOf > -1)
        count++;
    }

    return count;
  }
}

电话:

int occurrences = CharacterCounter.countOccurrences("l", "Hello World.");
System.out.println(occurrences); // 3
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); } }