我试图解决一个leetcode问题，但我的解决方案不是为某个测试用例返回正确的值。我想为这个解决方案实现两个指针技术（如果我使用了错误的技术，欢迎提供建议来解释正确的方法）。详情如下：
leetcode问题标题：
没有重复字符的最长子字符串。
问题：
给定一个字符串，求最长子字符串的长度，不要重复字符返回int）
我的解决方案：

public int lengthOfLongestSubstring(String s) {
        //place holder for me to create a "sub-string"
        String sub = "";

        //counter variable to count the characters in the substring
        int count = 0;
        int maxCount = 0;

        //my attempt at a TWO POINTER TECHNIQUE
        //pointers - to determine when the loop should break
        int head = 0;
        int tail = s.length();

        //this variable was intended to be used with the "indexOf" method, as the "start from" index...
        int index = 0;

        while(head < tail){
            //check if the next character in the string was already added to my substring.
            int val = sub.indexOf(s.charAt(head), index);

            //we proceed if it is -1 because this means the substring didnt previously contain that character
            if(val == -1){
                //added character to my substring
                sub+= s.charAt(head);
                count++;
                head++;
            } else {
                //reset data to default conditions, while continuing at the "head index" and check the rest of the substring
                count = 0;
                sub =  "";
            }
            //determine what the length of the longest substring.
            maxCount = count > maxCount ? count : maxCount;
        }
        return maxCount;
    }

我通过了测试用例：

> "" = expect 0
> " " = expect 1
> "abcabcbb" = expect 3
> "bbbbb" = expect 1
> "pwwkew" = expect 3
> "aab" = expect 2

我没有通过测试用例：

> "dvdgd" = expect 3, but got 2
> "dvjgdeds" = expect 5, but got 4
> "ddvbgdaeds" = expect 6, but got 4

可能的问题：
我相信发生这个问题是因为它用“v”传递索引，然后处理子字符串“dg”。因此，我试图更改解决方案，但修复该问题会导致我的所有其他案例返回错误，因此我认为应该放弃修复尝试。
我也尝试过：
我还尝试操作“indexof”方法来更改开始索引，只要在字符串中找到字符。然而，这产生了一个无限循环。
我已经试着解决这个问题好几个小时了，结果我累坏了。所以任何帮助都将不胜感激。如果可以，请详细解释，我对dsa和编程非常陌生，非常感谢。如果我需要更多的信息，请让我知道我会尽力回答。