I am trying to find duplicate segments in a Java array that was sorted by Arrays.sort(). I am expecting the same ints to form a duplicate segment in the array. For example, after sorting, the array is: {1, 2, 3, 3, 3, 3, 5, 6, 8, 8, 8, 8, 8, 9, 9, 9, 9}
I want to implement my following idea to find the duplicate segments: I want to use a while-loop with two pointers (i, and j).
1.) Let i start from the index 0, and let j start the last index (N-1). 2.) Keep i at the index 0 while doing j--. When j reaches the next index of i and no segments found, increment i by 1, and reinitialize j to index N-1 3. ) repeat steps 1 and 2. If a segment is found, increment i to j's current index, and reinitialize j to the index N-1. 4.) exit the while loop when i==j.
Below is my attempt, but it does not from my execution.
int[] test = new int[] {1, 2, 3, 3, 3, 3, 5, 6, 8, 8, 8, 8, 8, 9, 9, 9, 9};
int i = 0;
int j = test.length - 1;
int[] copySegment;
while (j > i)
{
if (test[j] == test[i])
{
int segmentLength = j - i + 1;
copySegment = new int[segmentLength];
for (int k = j; k >= i; k--)
{
copySegment[segmentLength--] = test[k];
}
for (int e : copySegment)
{
System.out.print(e + " ");
}
}
j--;
i++;
}
if you are trying to find the duplicate variables in your array and print the duplicates, may be you should start both the indexes from the beginning that is i=0, j=i+1
Arrays.sort(array); //sorts the array elements in ascending order
int i,j,lastNum;
ArrayList<Integer> list = new ArrayList<Integer>();
for(i=0,j=i+1;i<array.length-1;i++,j++) {
if(!list.isEmpty()) {
lastNum = list.get(list.size()-1);
} else {
lastNum = array[i]-1; // so that compiler doesn't warn you about lastNum not being initialized
}
if(array[i]!=lastNum) {
if(array[i]==array[j]) {
list.add(array[i]);
}
} else {
continue;
}
}
Iterator<Integer> it = list.iterator();
while(it.hasNext()) {
System.out.println(it.next().intValue());
}
EDIT: I have edited the answer to give a clear Explanation as @javaBeginner pointed. But this is unnecessary as the question clearly states that the array was sorted using Arrays.sort but anyways a clearer explanation won't hurt.