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Why the HashMap loading factor must be 0.75 instead of 0.8 force 0.6, many novices are not very clear about this, in order to help you solve this problem, the following editor will explain in detail for you, people with this need can come to learn, I hope you can get something.

Why does HashMap need a load factor?

The underlying layer of HashMap is the hash table, which is the type of structure that stores key-value pairs. It needs some calculation to determine where the data is stored in the hash table:

Static final int hash (Object key) {int h; return (key = = null)? 0: (h = key.hashCode ()) ^ (h > 16);} / / AbstractMappublic int hashCode () {int h = 0; Iterator I = entrySet (). Iterator (); while (i.hasNext ()) h + = i.next (). HashCode (); return h;}

The general data structure is either fast to query or fast to insert. HashMap is a data structure with slow insertion and fast query.

However, this data structure is prone to two problems: ①, if the space utilization is high, then when the hash algorithm calculates the storage location, you will find that many storage locations already have data (hash conflict); if ② increases the array capacity in order to avoid hash conflicts, it will lead to low space utilization.

The load factor represents the degree of filling of the elements in the Hash table.

Load factor = number of elements filled in the table / length of hash table

The larger the loading factor is, the more elements are filled, and the higher the space utilization is, but the chance of conflict is greater.

The smaller the loading factor is, the fewer elements are filled, the chance of conflict is reduced, but more space is wasted, and the number of expansion rehash operations is increased.

The greater the chance of conflict, the more expensive it is to find the data you need to find. Therefore, it is necessary to find a balance and compromise between "opportunity for conflict" and "space utilization".

So we can also know that there are several main factors that affect the efficiency of search:

Can the hash function hash the data in the hash table evenly?

How to deal with conflict?

How to choose the loading factor of the hash table?

This paper mainly introduces the latter two problems.

Is there any way to resolve the conflict? 1. The open addressing method Hi = (H (key) + di) MOD m, in which iLim 1 is 2, … , k (k

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