Merge sort on linkedlist
http://javabypatel.blogspot.in/2015/12/merge-sort-linked-list.html
https://www.cs.cmu.edu/~adamchik/15-121/lectures/Sorting%20Algorithms/code/MergeSort.java
X.
http://www.vogella.com/tutorials/JavaAlgorithmsMergesort/article.html
http://www.java2novice.com/java-sorting-algorithms/merge-sort/
X.
http://javahungry.blogspot.com/2013/06/java-sorting-program-code-merge-sort.html
It's stable.
http://www.vogella.com/tutorials/JavaAlgorithmsMergesort/article.html
public static void sort(Comparable[] a) {
Comparable[] aux = new Comparable[a.length];
sort(a, aux, 0, a.length-1);
assert isSorted(a);
}
http://www.sinbadsoft.com/blog/sorting-big-files-using-k-way-merge-sort/
It consists of two steps: first, split the file into small chunks that would fit in memory, load each chunk, sort it, and write it back on disk. Second, perform a k-way merge on all the sorted chunks to get the final result.
External Merge Sort
https://code.google.com/p/externalsortinginjava/source/browse/trunk/src/main/java/com/google/code/externalsorting/ExternalSort.java
http://exceptional-code.blogspot.com/2011/07/external-sorting-for-sorting-large.html
1. Start reading the input file from the beginning.
2. Read M (or less if number of entries remaining in the file is less than M) numbers from the file and store it into a temp buffer.
3. Sort (using any good sorting method - Quicksort, for example) the numbers in the buffer stored in step 2.
4. Create a temp file in disk and write out the sorted numbers from step 3 to this temp file. Save the name of the temp file.
5. Repeat step 2 to 4 until all numbers from the input file has been read, sorted, and written out to temp files.
6. Open all the temp files (and set the read pointer to the beginning of the files).
7. Find the minimum number from the set of numbers currently pointed to by the file read pointer.
8. Write the number to disk. (To increase efficiency you could write the number to a buffer first and then flush the buffer out to disk when the buffer is full. But modern I/O libraries should be doing this anyway for you).
9. Read another number from the file that contained the minimum number at step 7.
10. Repeat step 7 to 9 until all numbers from all the temp files have been processed, merged, and written out to disk
Mergesort using Fork/Join Framework
http://www.java-allandsundry.com/2012/08/mergesort-using-forkjoin-framework.html
http://www.oracle.com/technetwork/articles/java/fork-join-422606.html
https://github.com/cowtowncoder/java-merge-sort
http://www.geeksforgeeks.org/3-way-merge-sort/
Time Complexity: In case of 2-way Merge sort we get the equation: T(n) = 2T(n/2) + O(n)
Similarly, in case of 3-way Merge sort we get the equation: T(n) = 3T(n/3) + O(n)
By solving it using Master method, we get its complexity as O(n log 3n).. Although time complexity looks less compared to 2 way merge sort, the time taken actually may become higher because number of comparisons in merge function go higher
http://www.geeksforgeeks.org/binary-search-preferred-ternary-search/
http://javabypatel.blogspot.in/2015/12/merge-sort-linked-list.html
private Node mergeSortLinkList(Node startNode){ //Break the list until list is null or only 1 element is present in List. if(startNode==null || startNode.getNext()==null){ return startNode; } //Break the linklist into 2 list. //Finding Middle node and then breaking the Linled list in 2 parts. //Now 2 list are, 1st list from start to middle and 2nd list from middle+1 to last. Node middle = getMiddle(startNode); Node nextOfMiddle = middle.getNext(); middle.setNext(null); //Again breaking the List until there is only 1 element in each list. Node left = mergeSortLinkList(startNode); Node right = mergeSortLinkList(nextOfMiddle); //Once complete list is divided and contains only single element, //Start merging left and right half by sorting them and passing Sorted list further. Node sortedList = mergeTwoListRecursive(left, right); return sortedList; } //Recursive Approach for Merging Two Sorted List private Node mergeTwoListRecursive(Node leftStart, Node rightStart){ if(leftStart==null) return rightStart; if(rightStart==null) return leftStart; Node temp=null; if(leftStart.getData()<rightStart.getData()){ temp=leftStart; temp.setNext(mergeTwoListRecursive(leftStart.getNext(), rightStart)); }else{ temp=rightStart; temp.setNext(mergeTwoListRecursive(leftStart, rightStart.getNext())); } return temp; } private Node getMiddle(Node startNode) { if(startNode==null){ return startNode; } Node pointer1=startNode; Node pointer2=startNode; while(pointer2!=null && pointer2.getNext()!=null && pointer2.getNext().getNext()!=null){ pointer1 = pointer1.getNext(); pointer2 = pointer2.getNext().getNext(); } return pointer1; } //Iterative Approach for Merging Two Sorted List private Node mergeTwoListIterative(Node leftStart, Node rightStart) { Node merged=null; Node temp=null; //To keep track of last element, so that we don't need to iterate for adding the element at last of //list when either LeftStart or rightStart is NULL. Node lastAddedNode = null; while(leftStart!=null && rightStart!=null){ if(leftStart.getData()>rightStart.getData()){ temp = new Node(rightStart.getData()); rightStart=rightStart.getNext(); }else{ temp = new Node(leftStart.getData()); leftStart=leftStart.getNext(); } if(merged==null){ merged=temp; }else{ lastAddedNode.setNext(temp); } lastAddedNode=temp; } if(leftStart!=null){ lastAddedNode.setNext(leftStart); }else{ lastAddedNode.setNext(rightStart); } return merged; } public static void mergeSort(Comparable [ ] a)
{
Comparable[] tmp = new Comparable[a.length];
mergeSort(a, tmp, 0, a.length - 1);
}
private static void mergeSort(Comparable [ ] a, Comparable [ ] tmp, int left, int right)
{
if( left < right )
{
int center = (left + right) / 2;
mergeSort(a, tmp, left, center);
mergeSort(a, tmp, center + 1, right);
merge(a, tmp, left, center + 1, right);
}
}
private static void merge(Comparable[ ] a, Comparable[ ] tmp, int left, int right, int rightEnd )
{
int leftEnd = right - 1;
int k = left;
int num = rightEnd - left + 1;
while(left <= leftEnd && right <= rightEnd)
if(a[left].compareTo(a[right]) <= 0)
tmp[k++] = a[left++];
else
tmp[k++] = a[right++];
while(left <= leftEnd) // Copy rest of first half
tmp[k++] = a[left++];
while(right <= rightEnd) // Copy rest of right half
tmp[k++] = a[right++];
// Copy tmp back
for(int i = 0; i < num; i++, rightEnd--)
a[rightEnd] = tmp[rightEnd];
}
X.
http://www.vogella.com/tutorials/JavaAlgorithmsMergesort/article.html
http://www.java2novice.com/java-sorting-algorithms/merge-sort/
private int[] numbers;
private int[] helper;
private int number;
public void sort(int[] values) {
this.numbers = values;
number = values.length;
this.helper = new int[number];
mergesort(0, number - 1);
}
private void mergesort(int low, int high) {
// check if low is smaller then high, if not then the array is sorted
if (low < high) {
// Get the index of the element which is in the middle
int middle = low + (high - low) / 2;
// Sort the left side of the array
mergesort(low, middle);
// Sort the right side of the array
mergesort(middle + 1, high);
// Combine them both
merge(low, middle, high);
}
}
private void merge(int low, int middle, int high) {
// Copy both parts into the helper array
for (int i = low; i <= high; i++) {
helper[i] = numbers[i];
}
int i = low;
int j = middle + 1;
int k = low;
// Copy the smallest values from either the left or the right side back
// to the original array
while (i <= middle && j <= high) {
if (helper[i] <= helper[j]) {
numbers[k] = helper[i];
i++;
} else {
numbers[k] = helper[j];
j++;
}
k++;
}
// Copy the rest of the left side of the array into the target array
while (i <= middle) {
numbers[k] = helper[i];
k++;
i++;
}
}X.
http://javahungry.blogspot.com/2013/06/java-sorting-program-code-merge-sort.html
public static int[] mergeSort(int [] list) { if (list.length <= 1) { return list; } // Split the array in half int[] first = new int[list.length / 2]; int[] second = new int[list.length - first.length]; System.arraycopy(list, 0, first, 0, first.length); System.arraycopy(list, first.length, second, 0, second.length); // Sort each half mergeSort(first); mergeSort(second); // Merge the halves together, overwriting the original array merge(first, second, list); return list; } private static void merge(int[] first, int[] second, int [] result) { // Merge both halves into the result array // Next element to consider in the first array int iFirst = 0; // Next element to consider in the second array int iSecond = 0; // Next open position in the result int j = 0; // As long as neither iFirst nor iSecond is past the end, move the // smaller element into the result. while (iFirst < first.length && iSecond < second.length) { if (first[iFirst] < second[iSecond]) { result[j] = first[iFirst]; iFirst++; } else { result[j] = second[iSecond]; iSecond++; } j++; } // copy what's left System.arraycopy(first, iFirst, result, j, first.length - iFirst); System.arraycopy(second, iSecond, result, j, second.length - iSecond); }
It's stable.
http://www.vogella.com/tutorials/JavaAlgorithmsMergesort/article.html
public static void sort(Comparable[] a) {
Comparable[] aux = new Comparable[a.length];
sort(a, aux, 0, a.length-1);
assert isSorted(a);
}
http://www.sinbadsoft.com/blog/sorting-big-files-using-k-way-merge-sort/
It consists of two steps: first, split the file into small chunks that would fit in memory, load each chunk, sort it, and write it back on disk. Second, perform a k-way merge on all the sorted chunks to get the final result.
External Merge Sort
https://code.google.com/p/externalsortinginjava/source/browse/trunk/src/main/java/com/google/code/externalsorting/ExternalSort.java
http://exceptional-code.blogspot.com/2011/07/external-sorting-for-sorting-large.html
1. Start reading the input file from the beginning.
2. Read M (or less if number of entries remaining in the file is less than M) numbers from the file and store it into a temp buffer.
3. Sort (using any good sorting method - Quicksort, for example) the numbers in the buffer stored in step 2.
4. Create a temp file in disk and write out the sorted numbers from step 3 to this temp file. Save the name of the temp file.
5. Repeat step 2 to 4 until all numbers from the input file has been read, sorted, and written out to temp files.
6. Open all the temp files (and set the read pointer to the beginning of the files).
7. Find the minimum number from the set of numbers currently pointed to by the file read pointer.
8. Write the number to disk. (To increase efficiency you could write the number to a buffer first and then flush the buffer out to disk when the buffer is full. But modern I/O libraries should be doing this anyway for you).
9. Read another number from the file that contained the minimum number at step 7.
10. Repeat step 7 to 9 until all numbers from all the temp files have been processed, merged, and written out to disk
Mergesort using Fork/Join Framework
http://www.java-allandsundry.com/2012/08/mergesort-using-forkjoin-framework.html
http://www.oracle.com/technetwork/articles/java/fork-join-422606.html
https://github.com/cowtowncoder/java-merge-sort
http://www.geeksforgeeks.org/3-way-merge-sort/
public static void mergeSort3Way(Integer[] gArray) { // if array of size is zero returns null if (gArray == null) return; // creating duplicate of given array Integer[] fArray = new Integer[gArray.length]; // copying alements of given array into // duplicate array for (int i = 0; i < fArray.length; i++) fArray[i] = gArray[i]; // sort function mergeSort3WayRec(fArray, 0, gArray.length, gArray); // copy back elements of duplicate array // to given array for (int i = 0; i < fArray.length; i++) gArray[i] = fArray[i]; } /* Performing the merge sort algorithm on the given array of values in the rangeof indices [low, high). low is minimum index, high is maximum index (exclusive) */ public static void mergeSort3WayRec(Integer[] gArray, int low, int high, Integer[] destArray) { // If array size is 1 then do nothing if (high - low < 2) return; // Splitting array into 3 parts int mid1 = low + ((high - low) / 3); int mid2 = low + 2 * ((high - low) / 3) + 1; // Sorting 3 arrays recursively mergeSort3WayRec(destArray, low, mid1, gArray); mergeSort3WayRec(destArray, mid1, mid2, gArray); mergeSort3WayRec(destArray, mid2, high, gArray); // Merging the sorted arrays merge(destArray, low, mid1, mid2, high, gArray); } /* Merge the sorted ranges [low, mid1), [mid1, mid2) and [mid2, high) mid1 is first midpoint index in overall range to merge mid2 is second midpoint index in overall range to merge*/ public static void merge(Integer[] gArray, int low, int mid1, int mid2, int high, Integer[] destArray) { int i = low, j = mid1, k = mid2, l = low; // choose smaller of the smallest in the three ranges while ((i < mid1) && (j < mid2) && (k < high)) { if (gArray[i].compareTo(gArray[j]) < 0) { if (gArray[i].compareTo(gArray[k]) < 0) destArray[l++] = gArray[i++]; else destArray[l++] = gArray[k++]; } else { if (gArray[j].compareTo(gArray[k]) < 0) destArray[l++] = gArray[j++]; else destArray[l++] = gArray[k++]; } } // case where first and second ranges have // remaining values while ((i < mid1) && (j < mid2)) { if (gArray[i].compareTo(gArray[j]) < 0) destArray[l++] = gArray[i++]; else destArray[l++] = gArray[j++]; } // case where second and third ranges have // remaining values while ((j < mid2) && (k < high)) { if (gArray[j].compareTo(gArray[k]) < 0) destArray[l++] = gArray[j++]; else destArray[l++] = gArray[k++]; } // case where first and third ranges have // remaining values while ((i < mid1) && (k < high)) { if (gArray[i].compareTo(gArray[k]) < 0) destArray[l++] = gArray[i++]; else destArray[l++] = gArray[k++]; } // copy remaining values from the first range while (i < mid1) destArray[l++] = gArray[i++]; // copy remaining values from the second range while (j < mid2) destArray[l++] = gArray[j++]; // copy remaining values from the third range while (k < high) destArray[l++] = gArray[k++]; }Time Complexity: In case of 2-way Merge sort we get the equation: T(n) = 2T(n/2) + O(n)
Similarly, in case of 3-way Merge sort we get the equation: T(n) = 3T(n/3) + O(n)
By solving it using Master method, we get its complexity as O(n log 3n).. Although time complexity looks less compared to 2 way merge sort, the time taken actually may become higher because number of comparisons in merge function go higher
http://www.geeksforgeeks.org/binary-search-preferred-ternary-search/
int ternarySearch(int arr[], int l, int r, int x){ if (r >= l) { int mid1 = l + (r - l)/3; int mid2 = mid1 + (r - l)/3; // If x is present at the mid1 if (arr[mid1] == x) return mid1; // If x is present at the mid2 if (arr[mid2] == x) return mid2; // If x is present in left one-third if (arr[mid1] > x) return ternarySearch(arr, l, mid1-1, x); // If x is present in right one-third if (arr[mid2] < x) return ternarySearch(arr, mid2+1, r, x); // If x is present in middle one-third return ternarySearch(arr, mid1+1, mid2-1, x); } // We reach here when element is not present in array return -1;}
Which of the above two does less comparisons in worst case?
From the first look, it seems the ternary search does less number of comparisons as it makes Log3n recursive calls, but binary search makes Log2n recursive calls. Let us take a closer look.
The following is recursive formula for counting comparisons in worst case of Binary Search.
From the first look, it seems the ternary search does less number of comparisons as it makes Log3n recursive calls, but binary search makes Log2n recursive calls. Let us take a closer look.
The following is recursive formula for counting comparisons in worst case of Binary Search.
T(n) = T(n/2) + 2, T(1) = 1
The following is recursive formula for counting comparisons in worst case of Ternary Search.
T(n) = T(n/3) + 4, T(1) = 1
In binary search, there are 2Log2n + 1 comparisons in worst case. In ternary search, there are 4Log3n + 1 comparisons in worst case.
Time Complexity for Binary search = 2clog2n + O(1) Time Complexity for Ternary search = 4clog3n + O(1)
Therefore, the comparison of Ternary and Binary Searches boils down the comparison of expressions 2Log3n and Log2n . The value of 2Log3n can be written as (2 / Log23) * Log2n . Since the value of (2 / Log23) is more than one, Ternary Search does more comparisons than Binary Search in worst case.
