您可以为此目的使用Collectors.groupingBy(classifier,downstream) 方法:
List<Map<Integer, Integer>> listOfFreq = List.of(
// k1, v1, k2, v2, k3, v3, k4, v4, k5, v5, k6, v6
Map.of(47, 96, 82, 189, 100, 231, 125, 279, 158, 322, 240, 375),
Map.of(47, 125, 82, 239, 100, 285, 125, 334, 158, 378, 240, 429),
Map.of(47, 119, 82, 170, 100, 182, 125, 188, 158, 188, 240, 170));
Map<Integer, Integer> mapOfMaxFreq = listOfFreq
// Stream<Map<Integer,Integer>>
.stream()
// Stream<Set<Map.Entry<Integer,Integer>>>
.map(Map::entrySet)
// stream over all entries of all maps
// Stream<Map.Entry<Integer,Integer>>
.flatMap(Set::stream)
// looking for a max value for each key in
// all map entries and collect them to map
// Map<Integer,Optional<Map.Entry<Integer,Integer>>>
.collect(Collectors.groupingBy(
Map.Entry::getKey,
Collectors.maxBy(Comparator.comparingInt(Map.Entry::getValue))))
// Collection<Optional<Map.Entry<Integer,Integer>>>
.values()
// Stream<Optional<Map.Entry<Integer,Integer>>>
.stream()
// get value if present or else (0, 0)
// Stream<Map.Entry<Integer,Integer>>
.map(entry -> entry.orElse(Map.entry(0, 0)))
// collect to map
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
System.out.println(mapOfMaxFreq);
// {240=429, 82=239, 100=285, 125=334, 158=378, 47=125}