HashMap中的复合字符串键

问题描述

我们正在将一个String键存储在一个HashMap中，它是三个String字段和一个布尔字段的连接。问题是如果分隔符出现在字段值中，可以创建重复的键。

We are storing a String key in a HashMap that is a concatenation of three String fields and a boolean field. Problem is duplicate keys can be created if the delimiter appears in the field value.

所以要解决这个问题，根据发布，我打算创建一个将被用作HashMap的关键类键：

So to get around this, based on advice in another post, I'm planning on creating a key class which will be used as the HashMap key:

class TheKey {
  public final String k1;
  public final String k2;
  public final String k3;
  public final boolean k4;

  public TheKey(String k1, String k2, String k3, boolean k4) {
    this.k1 = k1; this.k2 = k2; this.k3 = k3; this.k4 = k4;
  }

  public boolean equals(Object o) {
      TheKey other = (TheKey) o;
      //return true if all four fields are equal
  }

  public int hashCode() {
    return ???;  
  }
}

我的问题是：

1. 从hashCode（）返回什么值。地图将共计约30个值。在这30个中，有大约10个不同的k1值（一些条目共享相同的k1值）。


2. 要将此密钥类存储为HashMap密钥，只需要覆盖equals（）和hashCode（）方法？还有其他需要吗？

推荐答案

只要hashCode和equals就可以了。 hashCode可能看起来像这样：

Just hashCode and equals should be fine. The hashCode could look something like this:

public int hashCode() {
  int hash = 17;
  hash = hash * 31 + k1.hashCode();
  hash = hash * 31 + k2.hashCode();
  hash = hash * 31 + k3.hashCode();
  hash = hash * 31 + k4 ? 0 : 1;
  return hash;
}

当然，这是假设没有任何键可以为空。通常，您可以使用0作为上述等式中的空引用的逻辑哈希码。需要处理空值的复合相等/哈希码的两个有用的方法：

That's assuming none of the keys can be null, of course. Typically you could use 0 as the "logical" hash code for a null reference in the above equation. Two useful methods for compound equality/hash code which needs to deal with nulls:

public static boolean equals(Object o1, Object o2) {
  if (o1 == o2) {
    return true;
  }
  if (o1 == null || o2 == null) {
    return false;
  }
  return o1.equals(o2);
}

public static boolean hashCode(Object o) {
  return o == null ? 0 : o.hashCode();
}

在此答案开始时，使用后一种方法在散列算法中，您'最终得到以下东西：

Using the latter method in the hash algorithm at the start of this answer, you'd end up with something like:

public int hashCode() {
  int hash = 17;
  hash = hash * 31 + ObjectUtil.hashCode(k1);
  hash = hash * 31 + ObjectUtil.hashCode(k2);
  hash = hash * 31 + ObjectUtil.hashCode(k3);
  hash = hash * 31 + k4 ? 0 : 1;
  return hash;
}

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