Approach#1
: Using two loops
==============================
In
this approach we Loop
through all elements in the array comparing an item with rest of the elements.
The outer loop will pick an item one by one and the
inner one compares the picked current element with the rest of the elements in
the array.
Here is the java implementation.
public static Object[]
RemoveDuplicatesV3(String[] toBeSearched){
List
toBeSearchedasList=Arrays.asList(toBeSearched);
List
toBeSearchedasListback= new ArrayList();
toBeSearchedasListback.addAll(toBeSearchedasList);
for(int
i=0;i
for(int
j=i+1;j
if(toBeSearchedasListback.get(i).equals(toBeSearchedasListback.get(j))){
toBeSearchedasListback.remove(j);
}
}
}
return
toBeSearchedasListback.toArray();
}
This involves O(n2) time. Where n is the no of
elements in the array.
Approach#2
: Sorting
==============================
First sort the array and traverse the array comparing each element with the next element and removing the element if it matches.
public static Object[]
RemoveDuplicatesV2(String[] toBeSearched){
List
toBeSearchedasList=Arrays.asList(toBeSearched);
List
toBeSearchedasListback= new ArrayList();
toBeSearchedasListback.addAll(toBeSearchedasList);
Collections.sort(toBeSearchedasListback);
for(int
i=0;i
System.out.println("in
loop");
if(i+1 !=
toBeSearchedasListback.size()
&&
toBeSearchedasListback.get(i).equals(toBeSearchedasListback.get(i+1))){
toBeSearchedasListback.remove(i);
}
}
return
toBeSearchedasListback.toArray();
}
This is better than approach#1
Approach#3
:Using HashMap
==============================
Put the element in hashmap while traversing. If it is already present remove it.
public static Object[]
RemoveDuplicates(String[] toBeSearched){
Hashtable original= new Hashtable();
for(String current
: toBeSearched){
if(original.get(current)!=null)
original.put(current,
current);
}
return
original.keySet().toArray();
}
This is best compared to first 2
