How to get different characters and their number in string in java

How to get different characters and their number in a string

This question can be broken down into two steps. The first step is to find the different characters. The second step is to find the different characters. Step 1, count their number. It seems a bit nonsense, right? Let me give you an answer first.

public class DistinctCharsCount {     public static void main(String[] args) {         printDistinctCharsWithCount("itwanger");         printDistinctCharsWithCount("chenmowanger");     }      private static void printDistinctCharsWithCount(String input) {         Map<Character, Integer> charsWithCountMap = new LinkedHashMap<>();          for (char c : input.toCharArray()) {             Integer oldValue = charsWithCountMap.get(c);              int newValue = (oldValue == null) ? 1 :                     Integer.sum(oldValue, 1);              charsWithCountMap.put(c, newValue);         }         System.out.println(charsWithCountMap);     } }

The output result of the program is:

{i=1, t=1, w=1, a=1, n=1, g=1, e=1, r=1} {c=1, h=1, e=2, n=2, m=1, o=1, w=1, a=1, g=1, r=1}

Let me talk about my ideas:

1) Declare a LinkedHashMap, or you can use HashMap, but the former can keep the string split After dividing the order, the results look more clear at a glance.

Why use Map? Because the key of Map is not allowed to be repeated, and it can just accumulate the number of repeated characters.

2) Split the string into characters and traverse.

3) If key is null, it means that its number is 1; otherwise, add 1 to the previous value, and then put it back into the Map, thus overwriting the previous number of characters.

The thinking is very clear, right? I can’t help but give myself a round of applause.

Well, after JDK 8, Map added a very powerful method merge(), which assigns values ​​to multiple keys at once:

private static void printDistinctCharsWithCountMerge(String input) {     Map<Character, Integer> charsWithCountMap = new LinkedHashMap<>();      for (char c : input.toCharArray()) {         charsWithCountMap.merge(c, 1, Integer::sum);     }     System.out.println(charsWithCountMap); }

Is it very powerful? It can be done with one line of code. The first parameter is the key, the second parameter is the value, and the third parameter is a BiFunction, which means that if the key already exists, recalculate the new value based on the BiFunction.

If the character appears for the first time, assign the value to 1; otherwise, sum the previous value to 1.

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