Ion*_*ona 5 java arraylist hashmap duplicates topmost
我试图在ArrayList中找到前10个最常见的字符串+它们的计数(出现频率).
我怎样才能以最佳时间复杂度做到这一点?
下面的代码在表单中找到最常见的单词+频率(String = int)
例如= 2
public static Entry<String, Integer> get10MostCommon(WordStream words) {
ArrayList<String> list = new ArrayList<String>();
Map<String, Integer> stringsCount = new HashMap<>();
Map.Entry<String, Integer> mostRepeated = null;
for (String i : words) {
list.add(i);
}
for (String s : list) {
Integer c = stringsCount.get(s);
if (c == null)
c = new Integer(0);
c++;
stringsCount.put(s, c);
}
for (Map.Entry<String, Integer> e : stringsCount.entrySet()) {
if (mostRepeated == null || mostRepeated.getValue() < e.getValue())
mostRepeated = e;
}
return mostRepeated;
}
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Fed*_*ner 13
您可以使用Java 8流分两步完成:
Map<String, Long> map = list.stream()
.collect(Collectors.groupingBy(w -> w, Collectors.counting()));
List<Map.Entry<String, Long>> result = map.entrySet().stream()
.sorted(Map.Entry.comparingByValue(Comparator.reverseOrder()))
.limit(10)
.collect(Collectors.toList());
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第一个流通过使用Collectors.groupingBy()with 来将单词映射到它们的频率Collectors.counting().
这将返回一个映射,其条目按照相反的顺序进行流式处理并按映射条目值排序.然后,流被限制为仅保留10个元素,这些元素最终被收集到列表中.