java - 计算字符串在字符串中出现的次数

Question

这是我对这种方法的尝试。

Count the number of co-occurrances of a non-empty sub-string sub within the string str E.g.

numOccurances("dogmonkeydog","dog") will return 2

numOccurances("dogmonkeydog","mon") will return 1

numOccurances("dogmonkeydog","cow") will return 0

public static int numOccurrences(String str, String sub) {
    int result = 0;
    int pos = str.indexOf(sub);
    if (pos == -1){
        return result;
    }
    if (sub.length() > str.length()){
        return result;
    }
    if ((str.substring(0, sub.length())).equals(sub)){
        result++;
        String st = str.substring(pos);
        return result + numOccurrences(st, sub); //Line 87
    }
    else{
        String st = str.substring(sub.length());
        return result + numOccurrences(st, sub);
    }
}

对于结果 > 0 的所有测试，我都会遇到此失败

java.lang.StackOverflowError
    at java.lang.String.indexOf(String.java:1718)
    at java.lang.String.indexOf(String.java:1698)
    at eecs2030.lab6.RecursiveTasks.numOccurrences(RecursiveTasks.java:77)
    at eecs2030.lab6.RecursiveTasks.numOccurrences(RecursiveTasks.java:87)

我不确定为什么我的代码从未达到其基本情况，任何见解将不胜感激!

Answer 1

您的方法中不需要第三个 if 条件。

 if ((str.substring(0, sub.length())).equals(sub))

您可以简单地将第三种情况定义为

 if(pos>=0)
  {
  result++;
  String newstr = str.substring(pos + sub.length());
  return numOccurrences(newstr,sub);
  }

因为，如果找到子字符串，pos 变量将通过子字符串的起始索引进行初始化，您可以在此处增加结果。

然后对字符串的其余部分递归调用 numOccurences() 方法。并在方法外部声明变量 result。

 import java.util.Scanner;
 class SubString
 {
 String user,subUser;
 Scanner sc=new Scanner(System.in);
 int num;
 static  int result = 0;
 int numOccurrences(String str, String sub) {

  int pos = str.indexOf(sub);
  if (pos == -1){
    return result;
 }
 if (sub.length() > str.length()){
    return result;
  }
  else if(pos >= 0)
  {
   result++;
   String newstr = str.substring(pos + sub.length());
   return numOccurrences(newstr,sub);
   }
 return result;

 }

//constructor
 SubString()
 {
   try{
   System.out.println("Enter string :");
   user=sc.nextLine();
   System.out.println("Enter the substring: ");
   subUser=sc.nextLine();
   num = numOccurrences(user,subUser);
   System.out.println(num);
  }
   catch(Exception e)
   {
    System.out.println(e);
   } 

 }
 public static void main(String...a)
{
 new SubString();
 }
 } 

`