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。

使用流和getOrDefault计数:

String s = "abcdeff";
s.chars().mapToObj(c -> (char) c)
 .forEach(c -> {
     int count = countMap.getOrDefault(c, 0) + 1;
     countMap.put(c, count);
  });

我不知道它有多高效，但下面的代码也可以工作。你需要在开头定义一个bifuncfunction。另外，你可以用这个方法做更多的增量。

public static Map<String, Integer> strInt = new HashMap<String, Integer>();

public static void main(String[] args) {
    BiFunction<Integer, Integer, Integer> bi = (x,y) -> {
        if(x == null)
            return y;
        return x+y;
    };
    strInt.put("abc", 0);


    strInt.merge("abc", 1, bi);
    strInt.merge("abc", 1, bi);
    strInt.merge("abc", 1, bi);
    strInt.merge("abcd", 1, bi);

    System.out.println(strInt.get("abc"));
    System.out.println(strInt.get("abcd"));
}

输出是

3
1

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.

你确定这是瓶颈吗?你做过性能分析吗?

尝试使用NetBeans分析器(它是免费的，内置在NB 6.1中)来查看热点。

最后，JVM升级(比如从1.5升级到>1.6)通常是一种廉价的性能增强。即使是版本号的升级也能提供良好的性能提升。如果您在Windows上运行，并且这是一个服务器类应用程序，请在命令行上使用-server来使用server Hotspot JVM。在Linux和Solaris机器上，这是自动检测到的。

函数Java库的TreeMap数据结构在最新的主干头中有一个更新方法:

public TreeMap<K, V> update(final K k, final F<V, V> f)

使用示例:

import static fj.data.TreeMap.empty;
import static fj.function.Integers.add;
import static fj.pre.Ord.stringOrd;
import fj.data.TreeMap;

public class TreeMap_Update
  {public static void main(String[] a)
    {TreeMap<String, Integer> map = empty(stringOrd);
     map = map.set("foo", 1);
     map = map.update("foo", add.f(1));
     System.out.println(map.get("foo").some());}}

这个程序输出“2”。