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Lecture 1 : (Learn 1D Linked List)

Creating a linked list node

class Node:
    def __init__(self,val=0,next=None):
        self.val = val
        self.next = next

node = Node(100)
print(node.val)
print(node.next)

100
None

Convert list to linked list

def to_linked_list(nums):
    if not nums:
        return None

    head = Node(nums[0])
    temp = head

    for num in nums[1:]:
        temp.next = Node(num)
        temp = temp.next

    return head

to_linked_list([1,2,3,4,5])

<__main__.Node at 0x107eb69d0>

Print 1D linked list

def from_linked_list(head):
    result = []
    temp = head

    while temp is not None:
        result.append(temp.val)
        temp = temp.next

    return result


ll = to_linked_list([1,2,3,4,5])
print(from_linked_list(ll))

[1, 2, 3, 4, 5]

Insert at the head of a Linked List

Given a linked list and an integer value val, insert a new node with that value at the beginning (before the head) of the list and return the updated linked list.

Input Format: 0->1->2, val = 5
Result: 5->0->1->2

Input Format:12->5->8->7, val = 100
Result: 100->12->5->8->7

def insert_beg(ll,val):
    return Node(val,ll)

print(from_linked_list(insert_beg(to_linked_list([0,1,2]),5)))  
print(from_linked_list(insert_beg(to_linked_list([12,5,8,7]),100)))  

[5, 0, 1, 2]
[100, 12, 5, 8, 7]

Complexity

• O(1)

• O(1)

Length of linked list

Given the head of a linked list, print the length of the linked list.

Input Format: 0->1->2
Result =3

Input Format: 2->5->8->7
Result: 4

Approach 1 (Brute Force)

• linear traversal and keep track of count

def linked_list_length(head):
    count = 0
    temp = head
    while temp is not None:
        count +=1
        temp = temp.next

    return count

print(linked_list_length(to_linked_list([0,1,2])))
print(linked_list_length(to_linked_list([2,5,8,7])))

3
4

Complexity

• O(N) : N = number of elements in the linked list

• O(1)

Approach 2 (Using Recursion)

• length(None) = 0
• length(node) = 1 + length(node.next)

def linked_list_length(head):
    if head is None:
        return 0

    return 1 + linked_list_length(head.next)
    

print(linked_list_length(to_linked_list([0,1,2])))
print(linked_list_length(to_linked_list([2,5,8,7])))

3
4

Complexity

• O(N) : N = number of elements in LL

• O(N) : N = number of elements in LL

Search an element in a Linked List

Given the head of a linked list and an integer value, find out whether the integer is present in the linked list or not. Return true if it is present, or else return false.

Input Format: 0->1->2, val = 2
Result True

Input Format: 12->5->8->7, val = 6 
Result False

Approach 1 (Using Linear Search)

def check_val(head,val):
    temp = head
    while temp is not None:
        if temp.val == val:
            return True
        temp = temp.next

    return False

print(check_val(to_linked_list([0,1,2]),2) )
print(check_val(to_linked_list([12,5,8,7]),6) )

True
False

Complexity

• O(N) : N = number of elements in linked list

• O(1)