design a method in Java which receives 2D array and find the most repetitive value for each column -

i going design method in java receives 2d array , find repetitive value each column. output method 1 dimensional array contains repeated value each column in 2 d array.

it can summarised that,

• count repetitive values each column.
• save these values in 1 array each value in output array represent repeated values in 2 d array column

this code, start

 static int  choseaction(int[][] actions, int colnumber) {     int action = 0;     int c = 0;     int d = 0;     int n = 0;           (int = 0; < actions.length; i++) {         (int j = 0; j < actions[0].length; j++) {              if (actions[colnumber][i] == 1) {                 c = +1;             } else if (actions[colnumber][i] == -1) {                 d = +1;             }              else if (actions[colnumber][i] == 0) {                 n = +1;             }         }      }      action = actioncompare(c, d, n);      return action; }  static int actioncompare(int a, int b, int c) {      int r;      if ((a > b) && (a > c)) {          r = a;         system.out.println("\n cc ");      } else if ((b > a) && (b > c)) {          r = b;         system.out.println("\n dd ");      } else {          r = c;         system.out.println("\n nn ");      }      return r;  } 

my question , easier way ?

here approach answer here

use hashmap<integer, integer>

for multiple occurrences, increment corresponding value of integer key;

public int[] static getfrequencies(int[][] actions){

int[] output = new int[actions.length] map<integer, integer> map = new hashmap<integer, integer>(); for(int j = 0; j < actions.length; j++){     (int : actions[j]) {         integer count = map.get(i);         map.put(i, count != null ? count+1 : 0);     } 

then append number maximum frequency hash map output array:

    output[j] = collections.max(map.entryset(),         new comparator<map.entry<integer, integer>>() {         @override         public int compare(entry<integer, integer> o1, entry<integer,     integer> o2) {             return o1.getvalue().compareto(o2.getvalue());         }     }).getkey().intvalue(); } 

and return output @ end:

return output; 

that's it!