Given a linked list of 0s, 1s and 2s, sort it.
Examples:
Input: 1 -> 1 -> 2 -> 0 -> 2 -> 0 -> 1 -> NULL Output: 0 -> 0 -> 1 -> 1 -> 1 -> 2 -> 2 -> NULL Input: 1 -> 1 -> 2 -> 1 -> 0 -> NULL Output: 0 -> 1 -> 1 -> 1 -> 2 -> NULL
Source: Microsoft Interview | Set 1
Following steps can be used to sort the given linked list.
- Traverse the list and count the number of 0s, 1s, and 2s. Let the counts be n1, n2, and n3 respectively.
- Traverse the list again, fill the first n1 nodes with 0, then n2 nodes with 1, and finally n3 nodes with 2.
Below image is a dry run of the above approach:
Below is the implementation of the above approach:
Python
# Python program to sort a linked list # of 0, 1 and 2 class LinkedList(object): def __init__(self): # Head of list self.head = None # Linked list Node class Node(object): def __init__(self, d): self.data = d self.next = None def sortList(self): # Initialise count of 0 1 # and 2 as 0 count = [0, 0, 0] ptr = self.head # Count total number of '0', '1' # and '2' # count[0] will store total number # of '0's # count[1] will store total number # of '1's # count[2] will store total number # of '2's while ptr != None: count[ptr.data] += 1 ptr = ptr.next i = 0 ptr = self.head # Let say count[0] = n1, count[1] = n2 # and count[2] = n3 # now start traversing list from head node, # 1) fill the list with 0, till n1 > 0 # 2) fill the list with 1, till n2 > 0 # 3) fill the list with 2, till n3 > 0 while ptr != None: if count[i] == 0: i+=1 else: ptr.data = i count[i] -= 1 ptr = ptr.next # Utility functions # Inserts a new Node at front of # the list. def push(self, new_data): # 1 & 2: Allocate the Node & # Put in the data new_node = self.Node(new_data) # 3. Make next of new Node as head new_node.next = self.head # 4. Move the head to point to new Node self.head = new_node # Function to print linked list def printList(self): temp = self.head while temp != None: print str(temp.data), temp = temp.next print '' # Driver code llist = LinkedList() llist.push(0) llist.push(1) llist.push(0) llist.push(2) llist.push(1) llist.push(1) llist.push(2) llist.push(1) llist.push(2) print "Linked List before sorting"llist.printList() llist.sortList() print "Linked List after sorting"llist.printList() # This code is contributed by BHAVYA JAIN |
Output:
Linked List Before Sorting 2 1 2 1 1 2 0 1 0 Linked List After Sorting 0 0 1 1 1 1 2 2 2
Time Complexity: O(n) where n is the number of nodes in the linked list.
Auxiliary Space: O(1)
Please refer complete article on Sort a linked list of 0s, 1s and 2s for more details!
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