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.
# 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!