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Remove Nth Node From End of List in Python
Suppose we have a linked list. We have to remove the Nth node from the end of the list, then return its head. So if the list is like [1, 2, 3, 4, 5, 6] and n = 3, then the returned list will be [1, 2, 3, 5, 6].
To solve this, we will follow these steps −
- If there is no node after head, then return None
- front := head, back := head, counter := 0 and fount := false
- while counter <= n
- if front is not present, then set flag as true, and come out from the loop
- front := next of front, and increase counter by 1
- while front is present
- front := next of front
- back := next of back
- if flag is false, then
- temp := next of back
- next of back := next of temp
- next of temp := None
- otherwise head := next of head
- Return head
Example(Python)
Let us see the following implementation to get a better understanding −
class ListNode:
def __init__(self, data, next = None):
self.val = data
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(']')
class Solution(object):
def removeNthFromEnd(self, head, n):
if not head.next:
return None
front=head
back = head
counter = 0
flag = False
while counter<=n:
if(not front):
flag = True
break
front = front.next
counter+=1
while front:
front = front.next
back = back.next
if not flag:
temp = back.next
back.next = temp.next
temp.next = None
else:
head = head.next
return head
head = make_list([1,2,3,4,5,6])
ob1 = Solution()
print_list(ob1.removeNthFromEnd(head, 3))Input
[1,2,3,4,5,6] 3
Output
[1,2,3,5,6]
Updated on: 27-Apr-2020
499 Views
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