why doesnt fullfiel the third condition?

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int values=1; HashMap<String, Integer> result = new HashMap<String, Integer>(); for (int unique = 0; unique <list.size(); unique++) { //verifies if the sort list has another unique word if(unique ==list.size()-1&&list.get(unique).equals(list.get(unique - 1))|| unique <list.size()-1&&list.get(unique).equals(list.get(unique + 1))){ //if it is at the last possible position add key and value to result if(unique ==list.size()-1) result.put(list.get(unique), values); values++; } //verifies if the last word and actual word are different else if(unique ==list.size()-1&&!list.get(unique).equals(list.get(unique - 1))|| unique <list.size()-1&&!list.get(unique).equals(list.get(unique + 1))){ //if they are different and value is greather than 1... return result of the last unique word and value if(values!=1){ result.put(list.get(unique), values); }//reset value values=1; } } return result;

Read a list of 20 words from the keyboard.
You need to count how many times each word appears in the list.
The result should be represented as a Map<String, Integer>, where the key is a unique word, and the value is the number of times the word appears in the list.

Display the contents of the map.
The case (uppercase/lowercase) affects the results.

Requirements:

The countWords method must declare and initialize a HashMap<String, Integer>.
The countWords method must return the created map.
The countWords method should add keys that correspond to unique words, and values that correspond to how many times those words occur.
The program should display the resulting map.
The main method must call the countWords method.

package com.codegym.task.task10.task1016; import java.io.BufferedReader; import java.io.InputStreamReader; import java.util.ArrayList; import java.util.HashMap; import java.util.Map; /* Identical words in a list */ public class Solution { public static void main(String[] args) throws Exception { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); ArrayList<String> words = new ArrayList<String>(); for (int i = 0; i < 20; i++) { words.add(reader.readLine()); } // words.sort(null); Map<String, Integer> map = countWords(words); for (Map.Entry<String, Integer> pair : map.entrySet()) { System.out.println(pair.getKey() + " " + pair.getValue()); } } public static Map<String, Integer> countWords(ArrayList<String> list) { int values=1; HashMap<String, Integer> result = new HashMap<String, Integer>(); for (int unique = 0; unique <list.size(); unique++) { //verifies if the sort list has another unique word if(unique ==list.size()-1&&list.get(unique).equals(list.get(unique - 1))|| unique <list.size()-1&&list.get(unique).equals(list.get(unique + 1))){ //if it is at the last possible position add key and value to result if(unique ==list.size()-1) result.put(list.get(unique), values); values++; } //verifies if the last word and actual word are different else if(unique ==list.size()-1&&!list.get(unique).equals(list.get(unique - 1))|| unique <list.size()-1&&!list.get(unique).equals(list.get(unique + 1))){ //if they are different and value is greather than 1... return result of the last unique word and value if(values!=1){ result.put(list.get(unique), values); }//reset value values=1; } } return result; } }

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Serg

Level 17 , Minsk, Belarus

19 June 2019, 12:58

By the way in Java 8 added a new convenient method: Map.putIfAbsent(K key, V value)

Guadalupe Gagnon

Level 37 , Tampa, United States

7 January 2019, 01:24

Here is the solution:

public static void main(String[] args) throws Exception {
   BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

   HashMap<String, Integer> map = new HashMap<>();
   for (int i = 0; i < 20; i++) {
      String word = reader.readLine();
      if(map.containsKey(word)) map.replace(word,map.get(word)+1);
      else map.put(word,1);
   }

   for (Map.Entry<String, Integer> pair : map.entrySet())
      System.out.println(pair.getKey() + " " + pair.getValue());
}

Guadalupe Gagnon

Level 37 , Tampa, United States

7 January 2019, 01:27

lines 7 and 8 are the big ones. You can already check if a hashmap already contains a key, which returns as a boolean t/f to use in an if statement. These two lines say if the word is already in the map, then replace the value with the old value plus 1, otherwise add the word with a value of 1. Hashmaps are a great tool for counting unique keys.

qwpo

Level 33

20 May 2019, 05:49

With Collections.frequency is even simpler. :)