Linked Hash Map counter not properly incrementing-CodePudding

Trying to "compress" a string by replacing duplicate characters with numbers (For example, the string aabcccccaaa would become a2blc5a3). I tried to use a Linked HashMap to solve the problem because the input order needs to be preserved, but the counter I created doesn't seem to be incrementing properly. Any insight would be greatly appreciated.

public class StringCompression {
    
    public static void main(String[] args) {
        
        String s = "aabcccccaaa";
        System.out.println(compString(s));
        
    }
    
    public static String compString(String str) {
        
        LinkedHashMap <Character, Integer> alphabet = new LinkedHashMap<>();
        StringBuilder strbdr = new StringBuilder();     
        
        for(int i = 0; i < str.length(); i  ) {
            
            if(alphabet.containsKey(str.charAt(i))) {
                alphabet.put(str.charAt(i), alphabet.get(str.charAt(i)) 1);
            }
            
            alphabet.put(str.charAt(i), 1);
        }
        
//      System.out.println(alphabet.entrySet());
        
        for(var entry : alphabet.entrySet()) {
            strbdr.append(entry.getKey());
            strbdr.append(entry.getValue());
        }
        
        return strbdr.toString();
    }
}

CodePudding user response：

Problem 1

The line alphabet.put(str.charAt(i), 1) keeps reseting each value to 1 , you need to put it in a else

for (int i = 0; i < str.length(); i  ) {
    if (alphabet.containsKey(str.charAt(i))) {
        alphabet.put(str.charAt(i), alphabet.get(str.charAt(i))   1);
    } else {
        alphabet.put(str.charAt(i), 1);
    }
}

Problem 2

First fix leads to a5b1c5 as maps have unique keys, so you can't count the a at the beginning and the a at the end

Just keep track of the previous char seen, and a counter

public static String compString(String str) {
    StringBuilder sb = new StringBuilder();
    char prev = '\0';
    int count = 0;
    for (char letter : str.toCharArray()) {
        if (prev == '\0') {
            prev = letter;
        } else if (prev != letter) {
            sb.append(prev).append(count);
            count = 0;
            prev = letter;
        }
        count  = 1;
    }
    return sb.append(prev).append(count).toString();
}

CodePudding user response：

alphabet.put(str.charAt(i), 1);

is not in an else statement and is executed everytime