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Program to find folded list from a given linked list in Python
Suppose we have a linked list. We have to take the first half of the linked list and fold over the second half then merge the intersecting nodes by taking their sum. Finally, we have to return the resulting head of the linked list.
So, if the input is like [5,8,1,2,4,7,5], then the output will be [2, 5, 15, 10, ]
To solve this, we will follow these steps −
- temp := 0
- ptr := node
- while ptr is not null, do
- temp := temp + 1
- ptr := next of ptr
- t := quotient of temp / 2
- m := node
- stk := a new stack
- while t is non-zero, do
- push value of m into stk
- tmp := next of m
- next of m := null
- m := tmp
- t := t - 1
- node := m
- if temp is even, then
- m := next of m
- while m is not null, do
- value of m := value of m + stack top element.
- pop from stk
- m := next of m
- return node
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):
temp = 0
ptr = node
while ptr:
temp += 1
ptr = ptr.next
t = temp // 2
m = node
stk = []
while t:
stk.append(m.val)
tmp = m.next
m.next = None
m = tmp
t -= 1
node = m
if temp % 2 != 0:
m = m.next
while m:
m.val += stk.pop()
m = m.next
return node
ob = Solution()
head = make_list([5,8,1,2,4,7,5])
print_list(ob.solve(head))Input
[5,8,1,2,4,7,5]
Output
[2, 5, 15, 10, ]
Updated on: 19-Nov-2020
192 Views
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