Using 2 pointers
This method does not need to use any containers, and does not need to create the array space. It is recommended, but the position information between the array elements is lost.
- Sort the array first;
- Initialize two pointers p, q to traverse the sorted array
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public static void solution(int[] array){
Arrays.sort(array);
int p = 0, q = 0;
for (int i = 0; i < array.length - 1; i++) {
if (array[i] != array[i+1]) {
if (p < q) {
array[p+1] = array[q+1];
p ++;
q ++;
} else {
p = i + 1;
}
} else {
q = i + 1;
}
}
for (int i = 0; i <= p; i++) {
System.out.print(array[i] + " ");
}
System.out.println("length = " + (p+1));
}
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Using Set
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Set set = new HashSet<>();
for(int i = 0; i < array.length; ++i){
set.add(array[i]);
}
int m = 0;
for(Iterator iterator = set.iterator(); iterator.hasNext();){
array[m] = (int)iterator.next();
m++;
}
for(int j = 0; j < set.size(); ++j){
System.out.print(array[j] + " ");
}
System.out.println();
System.out.println("length = " + set.size());
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Using List
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List list = new ArrayList<>();
for(int i = 0; i < array.length; ++i){
if(!list.contains(array[i]))
list.add(array[i]);
}
for(int i = 0; i < list.size(); ++i){
array[i] = (int)list.get(i);
}
for(int i = 0; i < list.size(); ++i){
System.out.print(array[i] + " ");
}
System.out.println();
System.out.println("length = " + list.size());
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