Java. util. bitset (a method for storing massive data)

Java. util. bitset can be stored by bit.
One byte occupies 8 bits in the computer. Data in Java is stored at least by byte,
For example, an int occupies 4 bytes.
If you encounter a large amount of data, it will inevitably require a large storage space and memory.
Algorithms can be used to reduce the storage space and memory occupied by data.
Java. util. bitset provides such an algorithm.
For example, there are a bunch of numbers that need to be stored, source = [3, 5, 6, 9]
Int requires 4*4 bytes.
Java. util. bitset can save true/false.
If Java. util. bitset is used, it will be much less. The principle is:
1. First find the maximum value in the Data maxvalue = 9
2. Declare a bitset BS. Its size is maxvalue + 1 = 10.
3. Traverse data source. Set BS [source [I] to true.
The last value is:
(0 is false; 1 is true)
BS [,]
3, 5, 6, 9

The Int type requires a number of 32 characters in 4 bytes. Currently, only one digit is used!
Proportion 32: 1
This saves a lot of space.

 

 

Test example

package com;import java.util.BitSet;public class MainTestThree {/** * @param args */public static void main(String[] args) {BitSet bm=new BitSet();System.out.println(bm.isEmpty()+"--"+bm.size());bm.set(0);System.out.println(bm.isEmpty()+"--"+bm.size());bm.set(1);System.out.println(bm.isEmpty()+"--"+bm.size());System.out.println(bm.get(65));System.out.println(bm.isEmpty()+"--"+bm.size());bm.set(65);System.out.println(bm.isEmpty()+"--"+bm.size());}}

Output:
True -- 64
False -- 64
False -- 64
False
False -- 64
False -- 128
 
It indicates that the default constructor declares a 64-bit bitset with a value of false.
If the number of bits you want to use exceeds the default size, it will apply for 64 bits instead of an error.

package com;import java.util.BitSet;public class MianTestFour {/** * @param args */public static void main(String[] args) {BitSet bm1=new BitSet(7);System.out.println(bm1.isEmpty()+"--"+bm1.size());BitSet bm2=new BitSet(63);System.out.println(bm2.isEmpty()+"--"+bm2.size());BitSet bm3=new BitSet(65);System.out.println(bm3.isEmpty()+"--"+bm3.size());BitSet bm4=new BitSet(111);System.out.println(bm4.isEmpty()+"--"+bm4.size());}}

 

Output:
True -- 64
True -- 64
True -- 128
True -- 128

It means that all the bits you applied for are multiples of 64. That is to say, if you apply for a bit no more than 64, the value is calculated as 64.
If the number is 2, it is 128.

 

Package com; import Java. util. bitset; public class maintestfive {/*** @ Param ARGs */public static void main (string [] ARGs) {int [] Shu =, 25, 31, 28, 37}; bitset bm1 = new bitset (maintestfive. getmaxvalue (SHU); system. out. println ("bm1.size () --" + bm1.size (); maintestfive. putvalueintobitset (Shu, bm1); printbitset (bm1);} // All values are false at the beginning. You can skip this because falsepublic static void initbitset (bitset BS) {fo R (INT I = 0; I <BS. size (); I ++) {BS. set (I, false) ;}}// print public static void printbitset (bitset BS) {stringbuffer Buf = new stringbuffer (); Buf. append ("[\ n"); For (INT I = 0; I <BS. size (); I ++) {if (I <BS. size ()-1) {Buf. append (maintestfive. getbitto10 (BS. get (I) + ",");} else {Buf. append (maintestfive. getbitto10 (BS. get (I);} If (I + 1) % 8 = 0 & I! = 0) {Buf. append ("\ n") ;}} Buf. append ("]"); system. out. println (BUF. tostring ();} // find the maximum value of the data set public static int getmaxvalue (INT [] zu) {int temp = 0; temp = zu [0]; for (INT I = 0; I <zu. length; I ++) {If (temp <zu [I]) {temp = zu [I] ;}} system. out. println ("maxvalue:" + temp); Return temp;} // put the public static void putvalue1_bitset (INT [] Shu, bitset BS) {for (INT I = 0; I <Shu. length; I ++) {BS. set (Shu [I], true) ;}// true, false to 1, 0 to look good public static string getbitto10 (Boolean flag) {string a = ""; if (flag = true) {return "1" ;}else {return "0 ";}}}

 

Output:
Maxvalue: 42
Bm1.size () -- 64
[
0, 0, 1, 0, 0, 1, 0,
0, 0, 0, 0, 0, 0, 1,
0, 0, 1, 0, 0, 0, 0,
,
0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0
]

This completes the storage value and value.
Note that it filters repeated numbers. That is to say, if a number appears more than twice, it is recorded as 1.
The information about the number of occurrences is lost.