Suppose we have a list L and another value k. We have to swap kth node from start and kth node from end and return the final list at end.
So, if the input is like L = [1,5,6,7,1,6,3,9,12] k = 3, then the output will be [1,5,3,7,1,6,6,9,12], the 3rd node from start is 6 and from end is 3, so they are swapped.
To solve this, we will follow these steps −
- temp := L
- for i in range 0 to k-2, do
- temp := next of temp
- firstNode := temp
- secondNode := L
- while next of temp is not null, do
- secondNode := next of secondNode
- temp := next of temp
- swap value of firstNode and value of secondNode
- return L
Example
Let us see the following implementation to get 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(']')
def solve(L, k):
temp = L
for i in range(k-1):
temp = temp.next
firstNode = temp
secondNode = L
while temp.next:
secondNode = secondNode.next
temp = temp.next
firstNode.val , secondNode.val = secondNode.val , firstNode.val
return L
L = [1,5,6,7,1,6,3,9,12]
k = 3
print_list(solve(make_list(L), k))Input
[1,5,6,7,1,6,3,9,12], 3
Output
[1, 5, 3, 7, 1, 6, 6, 9, 12, ]