JDK是在一直在迭代更新的,很多我们熟悉的类也悄悄的添加了一些新的方法特性。比如我们最常用的HashMap。
今天给大家讲一下HashMap在JDK8中添加的两个新方法compute和merge,从而实现一行代码实现单词统计的功能。一起来看看吧。
JDK8为我们引入了很多非常非常有用新特性,比如Stream和lambda表达式,可以让我们的程序更加简洁。
如果我们需要统计一个数组中单词出现的次数该怎么做呢?
这里不是讲算法,所以可以直接使用HashMap:
public void countBefore8(){ Map<String,Integer> wordCount= new HashMap<>(); String[] wordArray= new String[]{"we","are","the","world","we"}; for(String word: wordArray){ //如果存在则加1,否则将值设置为1 if(wordCount.containsKey(word)) { wordCount.put(word, wordCount.get(word) + 1); }else{ wordCount.put(word, 1); } } }
基本上流程是上面样子的。我们对数组进行遍历,然后判断这个单词是否存在于hashMap中,如果存在则+1。
逻辑很简单,但是看起来有些臃肿。
别怕,我们有JDK8。
先看下JDK8中compute的定义:
default V compute(K key, BiFunction<? super K, ? super V, ? extends V> remappingFunction) { Objects.requireNonNull(remappingFunction); V oldValue = get(key); V newValue = remappingFunction.apply(key, oldValue); if (newValue == null) { // delete mapping if (oldValue != null || containsKey(key)) { // something to remove remove(key); return null; } else { // nothing to do. Leave things as they were. return null; } } else { // add or replace old mapping put(key, newValue); return newValue; } }
可以看到compute有第二个参数BiFunction,BiFunction就是一个函数,输入两个参数,返回一个参数。
BiFunction的两个参数分别是key和key所对应的oldValue。
可考虑到我们的单词统计,我们可以直接将oldValue+1 即可。所以使用compute,可以将方法改写为:
public void countAfter8WithCompute(){ Map<String,Integer> wordCount= new HashMap<>(); String[] wordArray= new String[]{"we","are","the","world","we"}; Arrays.asList(wordArray).forEach(word ->{ wordCount.putIfAbsent(word,0); wordCount.compute(word,(w,count)->count+1); }); }
当然,我们可以将putIfAbsent放到compute中:
public void countAfter8WithCompute2(){ Map<String,Integer> wordCount= new HashMap<>(); String[] wordArray= new String[]{"we","are","the","world","we"}; Arrays.asList(wordArray).forEach(word -> wordCount.compute(word,(w, count)->count == null ? 1 : count + 1)); }
一行代码就完成了。
再看看merge方法:
default V merge(K key, V value, BiFunction<? super V, ? super V, ? extends V> remappingFunction) { Objects.requireNonNull(remappingFunction); Objects.requireNonNull(value); V oldValue = get(key); V newValue = (oldValue == null) ? value : remappingFunction.apply(oldValue, value); if (newValue == null) { remove(key); } else { put(key, newValue); } return newValue; }
merge方法需要3个参数,第一个参数是key,第二个参数是key对应的oldValue为空的值,也就是为空的默认值,第三个参数是一个BiFunction参数。
不同的是BiFunction的第一个参数是oldValue,第二个参数是value。
生成newValue的逻辑是:如果oldValue不存在,则使用value。如果oldValue存在,则调用BiFunction对oldValue和Value进行合并。
我们可以写出相应的代码如下:
public void countAfter8WithMerge(){ Map<String,Integer> wordCount= new HashMap<>(); String[] wordArray= new String[]{"we","are","the","world","we"}; Arrays.asList(wordArray).forEach(word->wordCount.merge(word, 1, (oldCount, one) -> oldCount + one)); }
后面的函数可以用Integer::sum替代:
public void countAfter8WithMerge(){ Map<String,Integer> wordCount= new HashMap<>(); String[] wordArray= new String[]{"we","are","the","world","we"}; Arrays.asList(wordArray).forEach(word->wordCount.merge(word, 1, Integer::sum)); }
本文的例子 https://github.com/ddean2009/learn-java-base-9-to-20/tree/master/java-base