Writing a method in Java that looks through an ArrayList of objects (specifically "card" objects, that contain the number and suit of the card) that has been filled with random cards. The method returns a boolean value whether the ArrayList contains a duplicate.
So far I've tried a for loop within a for loop, but I didn't get very far.
Also, I need another method that returns the duplicate card, but I need to finish this first method first before I attempt that.
This is how you would generally find duplicates:
public static bool findDuplicates(List<Card> listContainingDuplicates) {
final Set<Card> set1 = new HashSet<Card>();
for (Card yourCard : listContainingDuplicates) {
if (!set1.add(yourCard)) {
return true;
}
}
return false;
}
Looking at this: http://docs.oracle.com/javase/7/docs/api/java/util/ArrayList.html you could just use indexOf and lastIndexOf.
In psuedo code:
myfunct(list, obj){
if(list.indexOf(obj) == list.lastIndexOf(obj))
return False;
else
return True; // Two or more ...
}
true. I like the idea itself. It does not require extra space, though could be a hog on time.If your algorithm is limited in space, sort the list and then find duplicated values in liner time. The total time complexity of this algorithm is O(nlogn).
public boolean hadDuplicatedCards(List<Card> cards) {
Collections.sort(cards);
for (int i = 0; i < cards.size() - 1; i++) {
if(cards.get(i).equals(cards.get(i+1))) {
return true;
}
}
return false;
}
Otherwise, new a HashSet, and add cards one by one, if an equal card already present in HashSet, add method will return true.
BTW, In the sort approach, your Card class need to implement Comparable interface, and don't forget to override the equals method.
In the HashSet approach, you only need to override the equals method. But be aware that the HashSet approach needs extra O(n) space.
Thanks for PM 77-1's comments.
Card needs to implement Comparable interface.HashSet approach we need just equals method, that we can reasonably expect a class to have. Do not need Comparable here.
removeAllpassing in the second list, if the size changes, you have duplicate items...