将2D CharArray转换为String并使用.charAt（）比较字符是否正确？

Question

So I have a char variable called "temp" which I'd like to compare to the element stored in "X" CharArray[X][Y] while I'm in a third for loop after the 2D array.

For example:

char temp;
temp = ' ';
String end;
end = "";
for (int i = 0; i < CharArray.length; i++){
        for (int m = 0; m < 2; m++){
            if (somethingY){
                if (somethingZ){
                    for (int j = 0; j < something.length; j++){
                        //something
                        temp = somethingX;
                        if (temp == String.valueOf(CharArray[i][m]).charAt(0)){
                            end = String.valueOf(CharArray[i][m]);
                            System.out.print(end);
                        }
                    }
                }
            }
        }
    }

I've tried printing "temp" where it says "temp = somethingX" and it prints just fine. But when I try to save the String into a String variable , it will not print the variable called "end".

According to this , it won't do anything if the object is something else, but "end" is a String .

So, what am I doing wrong?

EDIT: In case there's a confusion, "I'm trying to print "end", but I figured if temp == String.valueOf(CharArray[i][m]).charAt(0) is correct, so should "end"'s part.".

EDIT2: Defined "temp" for people...

EDIT3: I tried " end.equals(String.valueOf(CharArray[i][m])); ", but still nothing happens when I try to print it. I get no errors nor anything.

EDIT4: I tried putting String.valueOf(CharArray[i][m]).charAt(0) into a another variable called "temp2" and doing if (temp == temp2) , but still the same thing.

EDIT5: I tried temp == CharArray[0][m] and then end = CharArray[0][m] , but still nothing prints.

EDIT6: OK. Sense this will never get resolved, I'll just say the whole point of my problem. -> I have an ArrayList where each line is a combination of a letter, space and a number (eg "E 3"). I need to check if a letter is repeating and if it is, I need to sum the numbers from all repeating letters.

For example, if I have the following ArrayList :

Z 3
O 9
I 1
J 7
Z 7
K 2
O 2
I 8
K 8
J 1

I need the output to be:

Z 10
O 11
I 9
J 8
K 10

I didn't want people to do the whole thing for me, but it seems I've no choice, since I've wasted 2 days on this problem and I'm running out of time.

Answer 1

Use a map :

ArrayList<String> input=new ArrayList<String>();
input.add("O 2");
input.add("O 2");
Map<String, Integer> map= new HashMap<String, Integer>();
for (String s:input) {
     String[] splitted=s.split(" ");
     String letter=splitted[0];
     Integer number=Integer.parseInt(splitted[1]);
     Integer num=map.get(letter);
     if (num==null) {
         map.put(letter,number);
     }
     else {
         map.put(letter,number+num);
     }
}
for (Map.Entry<String, Integer> entry : map.entrySet()) {
     System.out.println(entry.getKey() + " " + Integer.toString(entry.getValue()));
}

---

Without using a map :

ArrayList<String> input=new ArrayList<String>();
input.add("O 2");
input.add("O 2");
ArrayList<String> letters=new ArrayList<String>();
ArrayList<Integer> numbers=new ArrayList<Integer>();
for (String s:input) {
     String[] splitted=s.split(" ");
     String letter=splitted[0];
     Integer number=Integer.parseInt(splitted[1]);
     int index=-1;
     boolean isthere=false;
     for (String l:letters) {
          index++;
          if (l.equals(letter)) {
              isthere=true; //BUGFIX
              break;
          }
     }
     if (isthere==false) { //BUGFIX
         letters.add(letter);
         numbers.add(number);
     }
     else {
         numbers.set(index,numbers.get(index)+number);
     }      
}
for (int i=0; i < letters.size(); i++) {
     System.out.println(letters.get(i));
     System.out.print(numbers.get(i));
}

---

Converting it back to have a nice output :

ArrayList<String> output=new ArrayList<String>();
for (int i=0; i < letters.size(); i++) {
    output.add(letters.get(i)+" "+Integer.toString(numbers.get(i));
}

Feel free to comment if you are having any questions.