Guava學習筆記-簡化List分組成Map的實現

###問題描述
將一個List按照某種條件分組
[1, 2, 3, 4]->{OddNumber=[1, 3], EvenNumber=[2, 4]}

public static String getKey(Integer input) {
    return input % 2 == 0 ? "EvenNumber" : "OddNumber";
}

###原來

Map<String, List<Integer>> group = Maps.newHashMap();
for (Integer integer : source) {
    String key = getKey(integer);
    if (!group.containsKey(key)) {
        group.put(key, Lists.<Integer>newArrayList());
    }
    group.get(key).add(integer);
}

###使用Guava

Map<String, List<Integer>> group = Multimaps.asMap(
    Multimaps.index(source, new Function<Integer, String>() {
        public String apply(Integer input) {
            return getKey(input);
        }
    })
);

或

Map<String, Collection<Integer>> group = Multimaps.index(source,
    new Function<Integer, String>() {
        public String apply(Integer input) {
            return getKey(input);
        }
    }).asMap();

只須要關注key時怎麼實現的就行