Suppose we have linked list, we also have two values i and j, we have to reverse the linked list from i to jth nodes. And finally return the updated list.
So, if the input is like [1,2,3,4,5,6,7,8,9] i = 2 j = 6, then the output will be [1, 2, 7, 6, 5, 4, 3, 8, 9, ]
To solve this, we will follow these steps:
- prev_head := create a linked list node with value same as null and that points to node
- prev := prev_head, curr := node
- iterate through all values from 0 to i, do
- prev := curr, curr := next of curr
- rev_before := prev, rev_end := curr
- Iterate through all values from 0 to (j - i), do
- tmp := next of curr
- next of curr := prev
- prev, curr := curr, tmp
- next of rev_before := prev, rev_end.next := curr
- return next of prev_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, i, j):
prev_head = ListNode(None, node)
prev, curr = prev_head, node
for _ in range(i):
prev, curr = curr, curr.next
rev_before, rev_end = prev, curr
for _ in range(j - i + 1):
tmp = curr.next
curr.next = prev
prev, curr = curr, tmp
rev_before.next, rev_end.next = prev, curr
return prev_head.next
ob = Solution()
head = make_list([1,2,3,4,5,6,7,8,9])
i = 2
j = 6
print_list(ob.solve(head, i, j))Input
[1,2,3,4,5,6,7,8,9], 2, 6
Output
[1, 2, 7, 6, 5, 4, 3, 8, 9, ]