Reputation: 45
I keep getting a stackoverflow error on my mergesort call in my sort method. Its probably a quick fix, but I can't think of anything at the moment. Please look over my code.
public class MergeSorter {
public static <T> void sort(Comparable<? extends T>[] items) {
if (items == null) {
throw new IllegalArgumentException("Item is null.");
}
mergeSort(items, 0, items.length - 1);
}
@SuppressWarnings("unchecked")
private static <T> void mergeSort(Comparable<? extends T>[] items, int begIndx, int endIndx) {
if(begIndx == endIndx) {
return;
}
if(items.length > 1) {
int midIndx = items.length / 2;
mergeSort(items, begIndx, midIndx);
mergeSort(items, midIndx + 1, endIndx);
merge(items, begIndx, midIndx, endIndx);
}
}
@SuppressWarnings("unchecked")
private static <T> void merge(Comparable<? extends T>[] array, int begIndx, int midIndx, int endIndx) {
int sizeOfLeft = midIndx - begIndx + 1;
int sizeOfRight = endIndx - midIndx;
/// change to generic later
@SuppressWarnings("unchecked")
T[] leftArr = (T[]) new Object[sizeOfLeft + 1];
@SuppressWarnings("unchecked")
T[] rightArr = (T[]) new Object[sizeOfRight + 1];
for (int i = 0; i <= sizeOfLeft; i++) {
leftArr[i] = (T) array[begIndx + i];
}
for (int j = 0; j <= sizeOfRight; j++) {
rightArr[j] = (T) array[midIndx + j + 1];
}
int i = 0;
int j = 0;
for (int k = begIndx; k <= endIndx; k++) {
if (i == sizeOfLeft) {
array[k] = (Comparable<? extends T>) rightArr[j++];
} else if(j == sizeOfRight) {
array[k] = (Comparable<? extends T>) leftArr[i++];
} else if(((Integer) leftArr[i]).compareTo((Integer)rightArr[j]) <= 0) {
array[k] = (Comparable<? extends T>) leftArr[i++];
} else {
array[k]=(Comparable<? extends T>) rightArr[j++];
}
}
}
}
If I give it an Integer array of 5 things
5, 24, 79, 8, 3;
Then I should get 3,5,8,24,79 back.
Upvotes: 1
Views: 86
Reputation: 39324
Your problem is you always base your mid-index on items.length, which will be the same every call.
The mid index should be half way between your begIndx and endIndx for that call, that's what matters.
Upvotes: 1