The ArrayList is simple growing array. When trying to add element, and the buffer size is exceeded, it is simply growing. So the initial size can be any positive value. The 1 would be too little. Even with a few elements we will have a few resize operations. The 100 would be a loss of … Read more
You can’t use a foreach statement for that. The foreach is using internally an iterator: The iterators returned by this class’s iterator and listIterator methods are fail-fast: if the list is structurally modified at any time after the iterator is created, in any way except through the iterator’s own remove or add methods, the iterator … Read more
LinkedList might allocate fewer entries, but those entries are astronomically more expensive than they’d be for ArrayList — enough that even the worst-case ArrayList is cheaper as far as memory is concerned. (FYI, I think you’ve got it wrong; ArrayList grows by 1.5x when it’s full, not 2x.) See e.g. https://github.com/DimitrisAndreou/memory-measurer/blob/master/ElementCostInDataStructures.txt : LinkedList consumes 24 … Read more
A Map is a map, or “associative array”. It has a key->value layout. A List is on the other hand a list, which is an ordered collection of elements. A more direct comparison would possibly be between Set and List: Both these hold values, where the list is explicitly ordered (you can get element # … Read more
I’m curious, what do you need it for? You should know that the capacity is not (as it may sound) an upper limit of how much you can put into the ArrayList. It’s a value representing how much data you can put into the list, without forcing it to reallocate it internal array. Basically, the … Read more
Use the method “toArray()” ArrayList<String> mStringList= new ArrayList<String>(); mStringList.add(“ann”); mStringList.add(“john”); Object[] mStringArray = mStringList.toArray(); for(int i = 0; i < mStringArray.length ; i++){ Log.d(“string is”,(String)mStringArray[i]); } or you can do it like this: (mentioned in other answers) ArrayList<String> mStringList= new ArrayList<String>(); mStringList.add(“ann”); mStringList.add(“john”); String[] mStringArray = new String[mStringList.size()]; mStringArray = mStringList.toArray(mStringArray); for(int i = 0; … Read more
here’s some code that might demonstrate how it works out: import java.util.ArrayList; class A { private Long id; private String name; A(Long id){ this.id = id; } @Override public boolean equals(Object v) { boolean retVal = false; if (v instanceof A){ A ptr = (A) v; retVal = ptr.id.longValue() == this.id; } return retVal; } … Read more
On the definition of “sameness” As Joachim noted, for most application, List.equals(Object o) definition works: Compares the specified object with this list for equality. Returns true if and only if the specified object is also a list, both lists have the same size, and all corresponding pairs of elements in the two lists are equal. … Read more
Try this… ArrayList<String> values=new ArrayList<String>(); HashSet<String> hashSet = new HashSet<String>(); hashSet.addAll(values); values.clear(); values.addAll(hashSet);
For me (using Java 1.6.0_26), the first snippet gives the same exception as the second one. The reason is that the Arrays.asList(..) method does only return a List, not necessarily an ArrayList. Because you don’t really know what kind (or implementation of) of List that method returns, your cast to ArrayList<String> is not safe. The … Read more