Python Program For QuickSort On Singly Linked List

QuickSort on Doubly Linked List is discussed here. QuickSort on Singly linked list was given as an exercise. The important things about implementation are, it changes pointers rather swapping data and time complexity is same as the implementation for Doubly Linked List.

In partition(), we consider last element as pivot. We traverse through the current list and if a node has value greater than pivot, we move it after tail. If the node has smaller value, we keep it at its current position.

In QuickSortRecur(), we first call partition() which places pivot at correct position and returns pivot. After pivot is placed at correct position, we find tail node of left side (list before pivot) and recur for left list. Finally, we recur for right list.

Python3

# Sort a linked list using quick sort
class Node:
    def __init__(self, val):
        self.data = val
        self.next = None

class QuickSortLinkedList:
    def __init__(self):
        self.head=None

    def addNode(self, data):
        if (self.head == None):
            self.head = Node(data)
            return

        curr = self.head
        while (curr.next != None):
            curr = curr.next

        newNode = Node(data)
        curr.next = newNode

    def printList(self, n):
        while (n != None):
            print(n.data, end = " ")
            n = n.next

    ''' Takes first and last node,but do not
        break any links in the whole linked list'''
    def partitionLast(self, start, end):
        if (start == end or
            start == None or end == None):
            return start

        pivot_prev = start
        curr = start
        pivot = end.data

        ''' Iterate till one before the end,
            no need to iterate till the end
            because the end is pivot '''
        while (start != end):
            if (start.data < pivot):

                # Keep tracks of last
                # modified item
                pivot_prev = curr
                temp = curr.data
                curr.data = start.data
                start.data = temp
                curr = curr.next
            start = start.next

        ''' Swap the position of curr i.e.
            next suitable index and pivot'''
        temp = curr.data
        curr.data = pivot
        end.data = temp

        ''' Return one previous to current
            because current is now pointing
            to pivot '''
        return pivot_prev

    def sort(self, start, end):
        if(start == None or
           start == end or start == end.next):
            return

        # Split list and partition recurse
        pivot_prev = self.partitionLast(start,
                                       end)
        self.sort(start, pivot_prev)

        ''' If pivot is picked and moved to
            the start, that means start and
            pivot is the same so pick from
            next of pivot '''
        if(pivot_prev != None and
           pivot_prev == start):
            self.sort(pivot_prev.next, end)

        # If pivot is in between of the list,
        # start from next of pivot, since we
        # have pivot_prev, so we move two nodes
        elif (pivot_prev != None and
              pivot_prev.next != None):
            self.sort(pivot_prev.next.next,
                      end)

# Driver code
if __name__ == "__main__":
    ll = QuickSortLinkedList()
    ll.addNode(30)
    ll.addNode(3)
    ll.addNode(4)
    ll.addNode(20)
    ll.addNode(5)

    n = ll.head
    while (n.next != None):
        n = n.next

    print("Linked List before sorting")
    ll.printList(ll.head)

    ll.sort(ll.head, n)

    print("Linked List after sorting");
    ll.printList(ll.head)
# This code is contributed by humpheykibet

Output:

Linked List before sorting 
30 3 4 20 5 
Linked List after sorting 
3 4 5 20 30 

Time Complexity: O(N * log N), It takes O(N²) time in the worst case and O(N log N) in the average or best case.

Please refer complete article on QuickSort on Singly Linked List for more details!