为什么称为bitmap?
bitmap不仅仅存储介质以及数据结构不同于hashmap,存储的key和value也不同。
bitmap的key是元素的index,value只有0或者1(具体结构见下文)。
Bit-map的基本思想就是用一个bit位来标记某个元素对应的Value,而Key即是该元素。由于采用了Bit为单位来存储数据,因此可以很大程度上节省存储空间。
举例:
key-value: bitmap[1] = 1、bitmap[2]=0
添加:使用1和key所在位的value进行 |(或)
删除:使用1和key所在位的value进行 &(与)
位于java.util包中
/*
* BitSets are packed into arrays of "words." Currently a word is
* a long, which consists of 64 bits, requiring 6 address bits.
* The choice of word size is determined purely by performance concerns.
* 采用long作为载体,long有8个byte,所以有一个long有64个bit,64这个数字需要6个bit承载
*/
private final static int ADDRESS_BITS_PER_WORD = 6;
// 每一个words里面的元素占有64位
private final static int BITS_PER_WORD = 1 << ADDRESS_BITS_PER_WORD;
private final static int BIT_INDEX_MASK = BITS_PER_WORD - 1;
/* Used to shift left or right for a partial word mask */
private static final long WORD_MASK = 0xffffffffffffffffL;
/**
* @serialField bits long[]
*
* The bits in this BitSet. The ith bit is stored in bits[i/64] at
* bit position i % 64 (where bit position 0 refers to the least
* significant bit and 63 refers to the most significant bit).
*/
private static final ObjectStreamField[] serialPersistentFields = {
new ObjectStreamField("bits", long[].class),
};
/**
* The internal field corresponding to the serialField "bits".
* bitset的数据载体
*/
private long[] words;
/**
* The number of words in the logical size of this BitSet.
* 表示数组中最多使用的元素个数,也就是最后一个不为 0 的元素的索引加 1;比如[0,4,0,0],数组长度为 4,但是最后一个不为 0 的元素是 1,所以 wordsInUse = 2
*/
private transient int wordsInUse = 0;
创建一个 BitSet
对象时,默认 words
的长度为 1,并且 words[0] = 0
。当然也可以用户给定一个具体的容量大小,如下代码:
/**
* BitSet.class
* 创建一个能存储给定数据索引的 BitSet
*/
public BitSet(int nbits) {
// 参数合法性判断
if (nbits < 0)
throw new NegativeArraySizeException("nbits < 0: " + nbits);
// 调用 initWords 方法初始化
initWords(nbits);
sizeIsSticky = true;
}
private void initWords(int nbits) {
words = new long[wordIndex(nbits-1) + 1];
}
// 得到 bitIndex 对应的 words 下标
private static int wordIndex(int bitIndex) {
return bitIndex >> ADDRESS_BITS_PER_WORD;
}
public void set(int bitIndex) {
// 参数合法性检验
if (bitIndex < 0)
throw new IndexOutOfBoundsException("bitIndex < 0: " + bitIndex);
// 得到对应的数组下标
int wordIndex = wordIndex(bitIndex);
// 是否要扩容
expandTo(wordIndex);
// 修改数据
words[wordIndex] |= (1L << bitIndex);
// 参数检查
checkInvariants();
}
private void expandTo(int wordIndex) {
int wordsRequired = wordIndex+1;
if (wordsInUse < wordsRequired) {
// 扩容
ensureCapacity(wordsRequired);
wordsInUse = wordsRequired;
}
}
private void ensureCapacity(int wordsRequired) {
if (words.length < wordsRequired) {
// Allocate larger of doubled size or required size
// 基本上是扩容两倍
int request = Math.max(2 * words.length, wordsRequired);
words = Arrays.copyOf(words, request);
sizeIsSticky = false;
}
}
注意这里的set(bitIndex)
是让二进制的位置为1,并不是让words数组的某一index为1.
扩容的逻辑是:如果需要的长度大于数组的两倍,则扩容到需要的长度。否则,扩容位数组的两倍。
public void clear(int bitIndex) {
//...
int wordIndex = wordIndex(bitIndex);
// 如果 wordIndex >= wordsInUse,说明该索引要么不存在,要么一定是 0 ,直接返回即可
if (wordIndex >= wordsInUse)
return;
words[wordIndex] &= ~(1L << bitIndex);
recalculateWordsInUse();
//...
}
// 修改完可能会引起 wordsInUse 的变化,所以还要调用 recalculateWordsInUse() 重新计算 wordsInUse:从后往前遍历直到遇到 words[i] != 0,修改 wordsInUse = i+1。
private void recalculateWordsInUse() {
int i;
for (i = wordsInUse-1; i >= 0; i--)
if (words[i] != 0)
break;
wordsInUse = i+1; // The new logical size
}
public boolean get(int bitIndex) {
if (bitIndex < 0)
throw new IndexOutOfBoundsException("bitIndex < 0: " + bitIndex);
checkInvariants();
int wordIndex = wordIndex(bitIndex);
return (wordIndex < wordsInUse)
&& ((words[wordIndex] & (1L << bitIndex)) != 0);
}
/**
* Returns the number of bits of space actually in use by this
* {@code BitSet} to represent bit values.
* The maximum element in the set is the size - 1st element.
*
* @return the number of bits currently in this bit set
*/
public int size() {
return words.length * BITS_PER_WORD;
}
/**
* Returns the "logical size" of this {@code BitSet}: the index of
* the highest set bit in the {@code BitSet} plus one. Returns zero
* if the {@code BitSet} contains no set bits.
* 最高非0位+1
*
* @return the logical size of this {@code BitSet}
* @since 1.2
*/
public int length() {
if (wordsInUse == 0)
return 0;
return BITS_PER_WORD * (wordsInUse - 1) +
(BITS_PER_WORD - Long.numberOfLeadingZeros(words[wordsInUse - 1]));
}
集合操作还是很常用的,具体不作说明了,自行去看源码。
本文就到这里,下一篇介绍它的高级应用。