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Program to delete n nodes after m nodes from a linked list in Python
Suppose we are given a linked list that has the start node as "head", and two integer numbers m and n. We have to traverse the list and delete some nodes such as the first m nodes are kept in the list and the next n nodes after the first m nodes are deleted. We perform this until we encounter the end of the linked list. We start from the head node, and the modified linked list is to be returned.
The linked list structure is given to us as −
Node value : <integer> next : <pointer to next node>
So, if the input is like elements = [1, 2, 3, 4, 5, 6, 7, 8], m = 3, n = 1, then the output will be [1, 2, 3, 5, 6, 7, ]
Every node after 3 nodes is deleted in this process, so in the end the linked list will look like below −
To solve this, we will follow these steps −
prev := head
curr := head
q := 0
p := 0
while curr is not null, do
q := q + 1
if q is same as m, then
for i in range 0 to n, do
if curr.next is not null, then
curr := next of curr
next of prev := next of curr
q := 0
prev := next of prev
curr := next of curr
return head
Example (Python)
Let us see the following implementation to get a better understanding −
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def make_list(elements):
head = ListNode(elements[0])
for element in elements[1:]:
ptr = head
while ptr.next:
ptr = ptr.next
ptr.next = ListNode(element)
return head
def print_list(head):
ptr = head
print('[', end = "")
while ptr:
print(ptr.val, end = ", ")
ptr = ptr.next
print(']')
def solve(head, m, n):
prev = curr = head
q = 0
p = 0
while curr:
q += 1
if q == m:
for i in range(n):
if curr.next is not None:
curr = curr.next
prev.next = curr.next
q = 0
prev = prev.next
curr = curr.next
return head
head = ListNode()
elements = [1, 2, 3, 4, 5, 6, 7, 8]
head = make_list(elements)
res = solve(head, 3, 1)
print_list(res)Input
[1, 2, 3, 4, 5, 6, 7, 8], 3, 1
Output
[1, 2, 3, 5, 6, 7,]
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