字符的链接列表-int indexOf（String）

Question

How would you approach creating the method int indexOf(String str) if creating your own StringBuilder class as a linked list of char 's. Would you loop through with a bunch of logic or would it be easier to parse the StringBuilder object and then compare each one with str? I've made a few attempts but not having luck sorting out the logic in cases such as below:

Ex. query:

b1 = new MyStringBuilder("who is whoing over in whoville");
String s1 = new String("who");
String s2 = new String("whoing");
String s3 = new String("whoville");
String s4 = new String("whoviller");
String s5 = new String("wacky");  

My Current Method:
I can get this to work in cases where its at the beginning or if its not found but identifying str in the middle of the string is not working.

public int indexOf(String str)
    {
        int index =-1; //index at which str is first found in linked list string of chars
        int count = 0; //num of matches
        int firstI = -1;
        int sI=0; // dynamic counter variable to allow str.length and for loop to interact
        CNode currNode = firstC;
        for (int i = 0; i < length; i++)
        {
            if (currNode.data == str.charAt(sI))
            {
                if (count < 1)
                    firstI = i;
                count++;
            }

            if (count == 0 && (sI == str.length()-1))
                sI=0;

            if (count == str.length())
            {
                index = firstI;
                break;
            }

            if (count > 0 && currNode.data != str.charAt(sI))
            {
                sI = 0;
                count = 0;
            }


            currNode = currNode.next; //increment
            sI++;
        }

        return index;
    } 

Answer 1

Assuming MyStringBuilder has toString() , you could implement in terms of that.

If you aren't allowed to do that, you need to do some looping through your list of chars to find the first char. If that matches does the next char match the second in the word you are looking for and so on.