Java sample for the number of iterations of an array element and the size of a Java string comparison

public static Map<String, Integer> arraySearch(int[] array,int max){
  //The result set
  Map<String, Integer> resultMap = new HashMap<String, Integer>();
  //Number of repetitions
  int maxCount = 0;
  //Number of repetitions to many
  int value = 0;

  try{
    //Initializes an array of data that holds the number of occurrences of each element
    int[] dataArray = new int[max+1];

    //Traversing the array to be looked for, with each element as subscript, directly locate the data array, and perform the +1 operation to indicate that it appears once
    for(int i : array){
      dataArray[i]++;
    }

    //Find the maximum value in the data array
    for(int i=0;i<dataArray.length;i++){
      if(dataArray[i]>maxCount){
        maxCount=dataArray[i];
        value=i;
      }
    }
  }catch (Exception e) {}

  resultMap.put("maxCount", maxCount);
  resultMap.put("value", value);

  return resultMap;
} 

public static int compareString(String first,String second){
  int result = 0;

  try{
    //Turn null empty
    first = first==null?"":first;
    second = second==null?"":second;

    //Record the length of the string in advance to avoid repeated reads
    int firstLength=first.length();
    int secondLength=second.length();

    //Handles special cases with empty strings
    if("".equals(first) || "".equals(second)){
      //Whoever long small
      result = secondLength-firstLength;
    }else{
      //Temporary space for storing the sum of ASCII codes
      int firstCount = 0;
      int secondCount = 0;
      //The smaller of the two Numbers by pure arithmetic is really bt
      int minLength = (secondLength*(firstLength/secondLength) + firstLength*(secondLength/firstLength))/(firstLength/secondLength + secondLength/firstLength);
      //Truncate bit by bit the shorter number of bits in two strings to prevent overbounds
      for(int i=0;i<minLength;i++){
        //ASCII and
        firstCount+=first.substring(i,i+1).getBytes()[0];
        secondCount+=second.substring(i,i+1).getBytes()[0];
        //The sum is not equal to the sum, which means that the magnitude has been compared
        if(firstCount!=secondCount){
          break;
        }
      }

      if(firstCount==secondCount){
        //Long length
        result = firstLength-secondLength;
      }else{
        //The sum is big and the sum is big
        result = firstCount-secondCount;
      }
    }
  }catch (Exception e) {}

  return result;
}