问题

我正在使用一个HashMap<String, Integer>来跟踪特定字符串的出现次数。我以单线程的方式执行以下操作：

   HashMap<String, Integer> count = new HashMap<>();
   // List<String> words = ...;
   for (String word : words) {
      if (!count.containsKey(word)) {
          count.put(word, 0);
      }
      count.put(word, count.get(word) + 1);
   }

对于相同的单词，是否有可能计数增加超过1，因为我同时在同一个键上执行put和get操作？也就是说，假设word = "hello"。最初，count.get(word) = 1。当我执行count.put(word, count.get(word) + 1)时，如果我再执行count.get(word)，我得到的不是2，而是3。

   HashMap<String, Integer> count = new HashMap<>();
   // List<String≥ words = ...;
   for (String word : words) {
      if (!count.containsKey(word)) {
          count.put(word, 0);
      }
      count.put(word, count.get(word) + 1);
   }

答案1

得分: 2

Map具有compute和merge方法，可实现对键的值的简短更新：

1. compute

for (String word : words) {
    count.compute(word, (w, prev) -> prev == null ? 1 : prev + 1);
}

2. merge

for (String word : words) {
    count.merge(word, 1, (prev, one) -> prev + one);
}

Lambda表达式(prev, one) -> prev + one实际上是一个接受两个int参数并返回它们之和的函数，因此可以用方法引用Integer::sum替代：

for (String word : words) {
    count.merge(word, 1, Integer::sum);
}

for (String word : words) {
    count.compute(word, (w, prev) -> prev == null ? 1 : prev + 1);
}

for (String word : words) {
    count.merge(word, 1, (prev, one) -> prev + one);
}

for (String word : words) {
    count.merge(word, 1, Integer::sum);
}

答案2

得分: 2

直接回答你的问题：count.put(word, count.get(word) + 1) 这个语句不可能使得值增加超过 1。尽管这两个方法调用在同一个语句中，但它们是按顺序执行的：get 首先执行，以找到传递给 put 的第二个参数。

你可以将缺失键的测试和初始化合并为一条语句：

count.putIfAbsent(word, 0);

这会在返回值之后方便地执行：

count.put(word, 1 + count.putIfAbsent(word, 0));

然而，也有一种已经将这两个操作结合起来的方法：

count.merge(word, 1, Integer::sum);

count.putIfAbsent(word, 0);

count.put(word, 1 + count.putIfAbsent(word, 0));

count.merge(word, 1, Integer::sum);

答案3

得分: 1

以下是翻译好的部分：

• "It absolutely safe to do it in a single thread." -> "在单线程中执行这个操作是绝对安全的。"
• "No, it's not possible that 'count increases by more than 1 because I am performing a put and get on the same key at the same time' because two operations never can happen at the same time with single-threaded execution." -> "不，'count 增加超过 1 是因为我同时在相同的键上执行 put 和 get 操作' 是不可能的，因为在单线程执行中不会同时发生两个操作。"
• "Code count.put(word, count.get(word) + 1); will execute commands in following order:" -> "代码 count.put(word, count.get(word) + 1); 会按照以下顺序执行命令："
• "Integer value1 = count.get(word);" -> "Integer value1 = count.get(word);"
• "int value2 = value1.intValue();" -> "int value2 = value1.intValue();"
• "int value3 = value2 + 1;" -> "int value3 = value2 + 1;"
• "Integer value4 = new Integer(value3);" -> "Integer value4 = new Integer(value3);"
• "count.put(word, value4);" -> "count.put(word, value4);"
• "By the way, your code will produce quite a lot of garbage and will be not very effective." -> "顺便说一下，你的代码会产生大量的垃圾并且不太有效率。"
• "This way is more effective:" -> "这种方式更加有效："
• "private static class CounterHolder{" -> "private static class CounterHolder{"
• "int value;" -> "int value;"
• "Map<String, CounterHolder> count = new HashMap<>();" -> "Map<String, CounterHolder> count = new HashMap<>();"
• "List<String> words = ..." -> "List words = ..."
• "for (String word : words) {" -> "for (String word : words) {"
• "CounterHolder holder;" -> "CounterHolder holder;"
• "if (count.containsKey(word)) {" -> "if (count.containsKey(word)) {"
• "holder = new CounterHolder();" -> "holder = new CounterHolder();"
• "} else {" -> "} else {"
• "holder = new CounterHolder();" -> "holder = new CounterHolder();"
• "count.put(word, holder);" -> "count.put(word, holder);"
• "++holder.value;" -> "++holder.value;"

Integer value1 = count.get(word);
int value2 = value1.intValue();
int value3 = value2 + 1;
Integer value4 = new Integer(value3);
count.put(word, value4);

private static class CounterHolder{
    int value;
}
Map<String, CounterHolder> count = new HashMap<>();
List<String> words = ...
for (String word : words) {
    CounterHolder holder;
    if (count.containsKey(word)) {
        holder = new CounterHolder();
    } else {
        holder = new CounterHolder();
        count.put(word, holder);
    }
    ++holder.value;
}