google给安卓开发者推出了新型的Map存储集合，ArrayMap和SparseArray等。我们已经分析过了HashMap的实现原理（HashMap去重原理和内部实现），通过key的hash值把它放到数组的一个位置上，然后这个位置上形成一个链表结构，通过key的equals来判断相同。那google推出的有什么不同的地方吗？今天就让我们来看一看SparseArray到底是怎么回事。
其实SparseArray可以理解成一个键为int的map。内部有一个key的int型数组，一个value的object型数组，两个数组通过角标来形成对应关系。
public class SparseArray implements Cloneable {
private static final Object DELETED = new Object();
private boolean mGarbage = false;
private int[] mKeys;
private Object[] mValues;
private int mSize;
/**
* Creates a new SparseArray containing no mappings.
*/
public SparseArray() {
this(10);
}
/**
* Creates a new SparseArray containing no mappings that will not
* require any additional memory allocation to store the specified
* number of mappings. If you supply an initial capacity of 0, the
* sparse array will be initialized with a light-weight representation
* not requiring any additional array allocations.
*/
public SparseArray(int initialCapacity) {
if (initialCapacity == 0) {
mKeys = EmptyArray.INT;
mValues = EmptyArray.OBJECT;
} else {
mValues = ArrayUtils.newUnpaddedObjectArray(initialCapacity);
mKeys = new int[mValues.length];
}
mSize = 0;
}
public void append(int key, E value) {
if (mSize != 0 && key <= mKeys[mSize - 1]) {
put(key, value);
return;
}
if (mGarbage && mSize >= mKeys.length) {
gc();
}
mKeys = GrowingArrayUtils.append(mKeys, mSize, key);
mValues = GrowingArrayUtils.append(mValues, mSize, value);
mSize++;
}
/**
* Primitive int version of {@link #append(Object[], int, Object)}.
*/
public static int[] append(int[] array, int currentSize, int element) {
assert currentSize <= array.length;
if (currentSize + 1 > array.length) {
int[] newArray = ArrayUtils.newUnpaddedIntArray(growSize(currentSize));
System.arraycopy(array, 0, newArray, 0, currentSize);
array = newArray;
}
array[currentSize] = element;
return array;
}
private void gc() {
// Log.e("SparseArray", "gc start with " + mSize);
int n = mSize;
int o = 0;
int[] keys = mKeys;
Object[] values = mValues;
for (int i = 0; i < n; i++) {
Object val = values[i];
if (val != DELETED) {
if (i != o) {
keys[o] = keys[i];
values[o] = val;
values[i] = null;
}
o++;
}
}
mGarbage = false;
mSize = o;
// Log.e("SparseArray", "gc end with " + mSize);
}
public void put(int key, E value) {
int i = ContainerHelpers.binarySearch(mKeys, mSize, key);
if (i >= 0) {
mValues[i] = value;
} else {
i = ~i;
if (i < mSize && mValues[i] == DELETED) {
mKeys[i] = key;
mValues[i] = value;
return;
}
if (mGarbage && mSize >= mKeys.length) {
gc();
// Search again because indices may have changed.
i = ~ContainerHelpers.binarySearch(mKeys, mSize, key);
}
mKeys = GrowingArrayUtils.insert(mKeys, mSize, i, key);
mValues = GrowingArrayUtils.insert(mValues, mSize, i, value);
mSize++;
}
}
public static T[] insert(T[] array, int currentSize, int index, T element) {
assert currentSize <= array.length;
if (currentSize + 1 <= array.length) {
System.arraycopy(array, index, array, index + 1, currentSize - index);
array[index] = element;
return array;
}
@SuppressWarnings("unchecked")
T[] newArray = ArrayUtils.newUnpaddedArray((Class)array.getClass().getComponentType(),
growSize(currentSize));
System.arraycopy(array, 0, newArray, 0, index);
newArray[index] = element;
System.arraycopy(array, index, newArray, index + 1, array.length - index);
return newArray;
}
get方法也比较简单，大家看下就行了，不再分析。
public E get(int key) {
return get(key, null);
}
/**
* Gets the Object mapped from the specified key, or the specified Object
* if no such mapping has been made.
*/
@SuppressWarnings("unchecked")
public E get(int key, E valueIfKeyNotFound) {
int i = ContainerHelpers.binarySearch(mKeys, mSize, key);
if (i < 0 || mValues[i] == DELETED) {
return valueIfKeyNotFound;
} else {
return (E) mValues[i];
}
}
indexOf方法，因为key是有序的，所以使用二分，values只能顺序遍历。
/**
* Returns the index for which {@link #keyAt} would return the
* specified key, or a negative number if the specified
* key is not mapped.
*/
public int indexOfKey(int key) {
if (mGarbage) {
gc();
}
return ContainerHelpers.binarySearch(mKeys, mSize, key);
}
/**
* Returns an index for which {@link #valueAt} would return the
* specified key, or a negative number if no keys map to the
* specified value.
* Beware that this is a linear search, unlike lookups by key,
* and that multiple keys can map to the same value and this will
* find only one of them.
*
Note also that unlike most collections' {@code indexOf} methods,
* this method compares values using {@code ==} rather than {@code equals}.
*/
public int indexOfValue(E value) {
if (mGarbage) {
gc();
}
for (int i = 0; i < mSize; i++)
if (mValues[i] == value)
return i;
return -1;
}
clear也比较简单。
public void clear() {
int n = mSize;
Object[] values = mValues;
for (int i = 0; i < n; i++) {
values[i] = null;
}
mSize = 0;
mGarbage = false;
}
最后一个比较重要的是remove，里边控制了这个 mGarbage，然后是我们前边说的，不是真正删除，只是修改value的值,同时修改了mGarbage。
/**
* Alias for {@link #delete(int)}.
*/
public void remove(int key) {
delete(key);
}
/**
* Removes the mapping at the specified index.
*
* For indices outside of the range 0...size()-1,
* the behavior is undefined.
*/
public void removeAt(int index) {
if (mValues[index] != DELETED) {
mValues[index] = DELETED;
mGarbage = true;
}
}
/**
* Remove a range of mappings as a batch.
*
* @param index Index to begin at
* @param size Number of mappings to remove
*
* For indices outside of the range 0...size()-1,
* the behavior is undefined.
*/
public void removeAtRange(int index, int size) {
final int end = Math.min(mSize, index + size);
for (int i = index; i < end; i++) {
removeAt(i);
}
}
/**
* Removes the mapping from the specified key, if there was any.
*/
public void delete(int key) {
int i = ContainerHelpers.binarySearch(mKeys, mSize, key);
if (i >= 0) {
if (mValues[i] != DELETED) {
mValues[i] = DELETED;
mGarbage = true;
}
}
}
获取size的时候因为它是伪删除，所以需要判断 mGarbage，需要的时候进行gc。
/**
* Returns the number of key-value mappings that this SparseArray
* currently stores.
*/
public int size() {
if (mGarbage) {
gc();
}
return mSize;
}
到此为止，我们把这种map分析完了，对应的还有SparseBooleanArray, SparseIntArray, SparseLongArray只不过它们的value类型确定了下来，不过他们删除时是真正的删除，其它跟SparseArray完全一样。
还有一组 LongSparseArray,LongSparseLongArray跟这一组完全类似。