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Program to remove last occurrence of a given target from a linked list in Python
Suppose we have a singly linked list, and another value called target, we have to remove the last occurrence of target in the given list.
So, if the input is like [5,4,2,6,5,2,3,2,4,5,4,7], target = 5, then the output will be [5, 4, 2, 6, 5, 2, 3, 2, 4, 4, 7, ]
To solve this, we will follow these steps −
- head := node
- k := null, prev := null
- found := False
- while node is not null, do
- if value of node is same as target, then
- found := True
- prev := k
- k := node
- node := next of node
- if value of node is same as target, then
- if found is False, then
- return head
- if prev is null, then
- return next of head
- nect of prev := next of the next of prev
- return head
Let us see the following implementation to get better understanding −
Example
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:
def solve(self, node, target):
head = node
k = None
prev = None
found = False
while node:
if node.val == target:
found = True
prev = k
k = node
node = node.next
if found == False:
return head
if not prev:
return head.next
prev.next = prev.next.next
return head
ob = Solution()
L = make_list([5,4,2,6,5,2,3,2,4,5,4,7])
target = 5
print_list(ob.solve(L, target))
Input
[5,4,2,6,5,2,3,2,4,5,4,7]
Output
[5, 4, 2, 6, 5, 2, 3, 2, 4, 4, 7, ]
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