你应该意识到你最初的尝试

int count = map.containsKey(word) ? map.get(word) : 0;

包含映射上两个可能代价高昂的操作，即containsKey和get。前者执行的操作可能与后者非常相似，因此您要做两次相同的工作!

如果查看Map的API，当Map不包含所请求的元素时，get操作通常返回null。

注意，这将得到一个像

map.put( key, map.get(key) + 1 );

dangerous, since it might yield NullPointerExceptions. You should check for a null first. Also note, and this is very important, that HashMaps can contain nulls by definition. So not every returned null says "there is no such element". In this respect, containsKey behaves differently from get in actually telling you whether there is such an element. Refer to the API for details. For your case, however, you might not want to distinguish between a stored null and "noSuchElement". If you don't want to permit nulls you might prefer a Hashtable. Using a wrapper library as was already proposed in other answers might be a better solution to manual treatment, depending on the complexity of your application. To complete the answer (and I forgot to put that in at first, thanks to the edit function!), the best way of doing it natively, is to get into a final variable, check for null and put it back in with a 1. The variable should be final because it's immutable anyway. The compiler might not need this hint, but its clearer that way. final HashMap map = generateRandomHashMap(); final Object key = fetchSomeKey(); final Integer i = map.get(key); if (i != null) { map.put(i + 1); } else { // do something } If you do not want to rely on autoboxing, you should say something like map.put(new Integer(1 + i.getValue())); instead.

在java 8中，简单易行的方法如下:

final ConcurrentMap<String, AtomicLong> map = new ConcurrentHashMap<String, AtomicLong>();
    map.computeIfAbsent("foo", key -> new AtomicLong(0)).incrementAndGet();

我将使用Apache Collections Lazy Map(将值初始化为0)，并使用Apache Lang中的MutableIntegers作为该映射中的值。

在您的方法中，最大的代价是必须搜索两次地图。在我这里，你只需要做一次。只需要获取值(如果没有，它将被初始化)并增加它。

与其调用containsKey()，不如直接调用map更快。获取并检查返回值是否为空。

    Integer count = map.get(word);
    if(count == null){
        count = 0;
    }
    map.put(word, count + 1);

可以使用Java 8提供的Map接口中的computeIfAbsent方法。

final Map<String,AtomicLong> map = new ConcurrentHashMap<>();
map.computeIfAbsent("A", k->new AtomicLong(0)).incrementAndGet();
map.computeIfAbsent("B", k->new AtomicLong(0)).incrementAndGet();
map.computeIfAbsent("A", k->new AtomicLong(0)).incrementAndGet(); //[A=2, B=1]

方法computeIfAbsent检查指定的键是否已经与某个值关联?如果没有关联值，则尝试使用给定的映射函数计算其值。在任何情况下，它都会返回与指定键关联的当前值(现有值或计算值)，如果计算值为空则返回null。

另一方面，如果你遇到多个线程更新一个公共和的情况，你可以看看LongAdder类。在高争用情况下，该类的预期吞吐量显著高于AtomicLong，但代价是更高的空间消耗。

查看谷歌Collections Library总是一个好主意。在这种情况下，Multiset将做的伎俩:

Multiset bag = Multisets.newHashMultiset();
String word = "foo";
bag.add(word);
bag.add(word);
System.out.println(bag.count(word)); // Prints 2

有类似map的方法用于遍历键/条目等。在内部实现目前使用HashMap<E, AtomicInteger>，所以您不会产生装箱成本。