Counting the number of elements in a List and appending it to the end of each item and maintaining insertion order in list to be returned?

huangapple
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英文:

Counting the number of elements in a List and appending it to the end of each item and maintaining insertion order in list to be returned?

问题

  1. public class DeviceManagement {
  2.     public static List<String> Solution(List<String> ls) {
  3.         Map<String, Integer> map = new LinkedHashMap<>();
  4.         int counter = 1;
  6.         List<String> result = new ArrayList<>();
  7.         for (String str : ls) {
  8.             if (map.containsKey(str)) {
  9.                 int frequency = map.get(str);
  10.                 map.put(str, frequency + 1);
  11.                 result.add(str + frequency);
  12.             } else {
  13.                 map.put(str, counter);
  14.                 result.add(str);
  15.             }
  16.         }
  18.         return result;
  19.     }
  21.     public static void main(String[] args) {
  22.         List<String> ls = new ArrayList<>();
  23.         ls.add("speaker");
  24.         ls.add("tv");
  25.         ls.add("radio");
  26.         ls.add("toaster");
  27.         ls.add("radio");
  28.         ls.add("speaker");
  29.         System.out.println(ls);
  30.         System.out.println(Solution(ls));
  31.     }
  32. }
英文:

I am working on a problem and needed some possible approaches to it.

Sample input:

  1. [&quot;tv&quot;, &quot;speaker&quot;, &quot;tv&quot;, &quot;radio&quot;, &quot;radio&quot;, &quot;tv&quot;]

Sample output:

  1. [&quot;tv&quot;, &quot;speaker&quot;, &quot;tv1&quot;, &quot;radio&quot;, &quot;radio1&quot;, &quot;tv2&quot;]

Add the number of occurrences while leaving the first occurrence as it is.

So far I have added the elements to a HashMap and using Collection.frequency/get put for the element count to count the number of elements. However, what would be a way to append the corresponding number at the end?

  1. public class DeviceManagement {
  2.     public static List&lt;String&gt; Solution(List&lt;String&gt; ls) {
  3. 	    Map&lt;String, Integer&gt; map = new LinkedHashMap&lt;&gt;();
  4. 	    int counter = 1;
  5. 	 
  6.         for (String str : ls) {
  7. 		    if (map.containsKey(str)) {
  8. 			    map.put(str, map.get(str) + 1);
  9. 		    } else {
  10.                 map.put(str, counter);
  11.             }
  12. 	    }
  13. 	    
  14.         System.out.println(map);
  15. 	    return null;
  16.     }
  18.     public static void main(String[] args) {
  19. 	    List&lt;String&gt; ls = new ArrayList&lt;&gt;();
  20. 	    ls.add(&quot;speaker&quot;);
  21. 	    ls.add(&quot;tv&quot;);
  22. 	    ls.add(&quot;radio&quot;);
  23. 	    ls.add(&quot;toaster&quot;);
  24. 	    ls.add(&quot;radio&quot;);
  25. 	    ls.add(&quot;speaker&quot;);
  26. 	    System.out.println(ls);
  27. 	    System.out.println(Solution(ls));
  28.     }
  29. }

答案1

得分: 0

您的代码正确地检测出重复项并计算频率，但在迭代过程中未替换 List 中的任何元素为其新值。为了简化此过程，我建议使用 List#replaceAll 替代 for 循环。您的代码将类似于以下内容：

  1. public static List<String> solution(List<String> ls) {
  2.     Map<String, Integer> map = new HashMap<>();
  4.     ls.replaceAll(element -> {
  5.         if (map.containsKey(element)) {
  6.             int oldAmount = map.put(element, map.get(element) + 1);
  7.             return element + oldAmount;
  8.         } else {
  9.             map.put(element, 1);
  10.             return element;
  11.         }
  12.     });
  14.     return ls;
  15. }

这将使您的输出为：

  1. [speaker, tv, radio, toaster, radio1, speaker1]

为了稍微简化代码，您可以使用 Map#merge：

  1. public static List<String> solution(List<String> ls) {
  2.     Map<String, Integer> map = new HashMap<>();
  4.     ls.replaceAll(element -> {
  5.         int newValue = map.merge(element, 1, Integer::sum);
  6.         return newValue == 1 ? element : element + (newValue - 1);
  7.     });
  9.     return ls;
  10. }

如果您想返回另一个不同的 List 对象，您可以对输入进行流式处理，映射它，并将其收集到另一个 List 中：

  1. public static List<String> solution(List<String> ls) {
  2.     Map<String, Integer> map = new HashMap<>();
  4.     return ls.stream().map(element -> {
  5.         int newValue = map.merge(element, 1, Integer::sum);
  6.         return newValue == 1 ? element : element + (newValue - 1);
  7.     }).collect(Collectors.toList());
  8. }
英文:

Your code correctly detects duplicates and counts frequencies, but doesn't replace any elements in the List with their new values while you're iterating over it. To make this easy, I recommend using List#replaceAll instead of a for-loop. Your code would look something like:

  1. public static List&lt;String&gt; solution(List&lt;String&gt; ls) {
  2.     Map&lt;String, Integer&gt; map = new HashMap&lt;&gt;();
  4.     ls.replaceAll(element -&gt; {
  5.         if (map.containsKey(element)) {
  6.             int oldAmount = map.put(element, map.get(element) + 1);
  7.             return element + oldAmount;
  8.         } else {
  9.             map.put(element, 1);
  10.             return element;
  11.         }
  12.     });
  14.     return ls;
  15. }

This would result in your output being:

  1. [speaker, tv, radio, toaster, radio1, speaker1]

To simplify your code a bit, you can utilize Map#merge:

  1. public static List&lt;String&gt; solution(List&lt;String&gt; ls) {
  2.     Map&lt;String, Integer&gt; map = new HashMap&lt;&gt;();
  4.     ls.replaceAll(element -&gt; {
  5.         int newValue = map.merge(element, 1, Integer::sum);
  6.         return newValue == 1 ? element : element + (newValue - 1);
  7.     });
  9.     return ls;
  10. }

If you want to return a different List object, then you can stream your input, map it, and collect it to another List:

  1. public static List&lt;String&gt; solution(List&lt;String&gt; ls) {
  2.     Map&lt;String, Integer&gt; map = new HashMap&lt;&gt;();
  4.     return ls.stream().map(element -&gt; {
  5.         int newValue = map.merge(element, 1, Integer::sum);
  6.         return newValue == 1 ? element : element + (newValue - 1);
  7.     }).collect(Collectors.toList());
  8. }

huangapple
  • 本文由 发表于 2020年3月15日 09:39:28
  • 转载请务必保留本文链接：https://go.coder-hub.com/60688954.html
  • algorithm
  • collections
  • hashmap
  • java
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