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[英]Elegant way to count how many pieces of each type are present in a position
[英]the quickest way to Implement an algorithm to count how many times each string is present in an array of strings using Java
只是想知道實現一種算法以使用Java計算字符串數組中每個字符串存在多少次的最快方法是什么?
這是我嘗試過的方法,並且有效,但是我擔心它可能會“作弊”,因為它偏離了這個問題:
{
String[] stringArray = {"Random", "Words", "Here","Random", "Words", "Here","Random", "Words", "Here","Random", "Words", "Here"};
List asList = Arrays.asList(stringArray);
Set<String> mySet = new HashSet<>(asList);
mySet.stream().forEach((s) -> {
System.out.println(s + " " +Collections.frequency(asList,s));
});
}
最簡單的方法是使用Map#merge()
:
Map<String, Integer> m = new HashMap<>();
for (String s : array)
m.merge(s, 1, Integer::sum);
之后, m
將把字符串作為鍵,將出現的值作為值:
m.forEach((k, v) -> System.out.format("%s occured %s time(s)\n", k, v));
通過在流中使用收集器:
Arrays.stream(list).collect(Collectors.groupingBy(e -> e, Collectors.counting()))
因此,如果您有這樣的事情:
String[] list = new String[4];
list[0] = "something";
list[1] = "gfddfgdfg";
list[2] = "something";
list[3] = "somet444hing";
System.out.println(Arrays.stream(list).collect(Collectors.groupingBy(e -> e, Collectors.counting())));
輸出將是:
{gfddfgdfg=1, something=2, somet444hing=1}
我將使用groupingBy返回計數圖。
Map<String, Long> counts = Stream.of(array)
.collect(Collectors.groupingBy(w -> w, Collectors.counting()));
也可以打印這些
Stream.of(array)
.collect(Collectors.groupingBy(w -> w, Collectors.counting()))
.forEach((k, v) -> System.out.println(k + " occurred " + v " times));
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