How does a Java HashMap handle different objects with the same hash code?

Question

As per my understanding I think:

1. It is perfectly legal for two objects to have the same hashcode.
2. If two objects are equal (using the equals() method)

Answer 1

Here is a rough description of HashMap's mechanism, for Java 8 version, (it might be slightly different from Java 6).

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Data structures

• Hash table
  Hash value is calculated via hash() on key, and it decide which bucket of the hashtable to use for a given key.
• Linked list (singly)
  When count of elements in a bucket is small, a singly linked list is used.
• Red-Black tree
  When count of elements in a bucket is large, a red-black tree is used.

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Classes (internal)

• Map.Entry
  Represent a single entity in map, the key/value entity.

• HashMap.Node
  Linked list version of node.

  It could represent:

  • A hash bucket.
    Because it has a hash property.
  • A node in singly linked list, (thus also head of linkedlist).
• HashMap.TreeNode
  Tree version of node.

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Fields (internal)

• Node[] table
  The bucket table, (head of the linked lists).
  If a bucket don't contains elements, then it's null, thus only take space of a reference.
• Set entrySet Set of entities.
• int size
  Number of entities.
• float loadFactor
  Indicate how full the hash table is allowed, before resizing.
• int threshold
  The next size at which to resize.
  Formula: threshold = capacity * loadFactor

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Methods (internal)

• int hash(key)
  Calculate hash by key.

• How to map hash to bucket?
  Use following logic:

  static int hashToBucket(int tableSize, int hash) {
      return (tableSize - 1) & hash;
  }
  

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About capacity

In hash table, capacity means the bucket count, it could be get from table.length.
Also could be calculated via threshold and loadFactor, thus no need to be defined as a class field.

Could get the effective capacity via: capacity()

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Operations

• Find entity by key.
  First find the bucket by hash value, then loop linked list or search sorted tree.
• Add entity with key.
  First find the bucket according to hash value of key.
  Then try find the value:
  • If found, replace the value.
  • Otherwise, add a new node at beginning of linked list, or insert into sorted tree.
• Resize
  When threshold reached, will double hashtable's capacity(table.length), then perform a re-hash on all elements to rebuild the table.
  This could be an expensive operation.

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Performance

• get & put
  Time complexity is O(1), because:
  • Bucket is accessed via array index, thus O(1).
  • Linked list in each bucket is of small length, thus could view as O(1).
  • Tree size is also limited, because will extend capacity & re-hash when element count increase, so could view it as O(1), not O(log N).