Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

Given linked list – head = [4,5,1,9], which looks like following:

Example 1:

Input: head = [4,5,1,9], node = 5

Output: [4,1,9]

Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function.

Example 2:

Input: head = [4,5,1,9], node = 1

Output: [4,5,9]

Explanation: You are given the third node with value 1, the linked list should become 4 -> 5 -> 9 after calling your function.

Note:

• The linked list will have at least two elements.
• All of the nodes’ values will be unique.
• The given node will not be the tail and it will always be a valid node of the linked list.
• Do not return anything from your function.

Solution

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def deleteNode(self, node: ListNode) -> None:
        if node.next:
            node.val = node.next.val
            node.next = node.next.next

Test Cases

test = Solution()
head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
head.next.next.next = ListNode(4)
test.deleteNode(head)
assert head.val == 2
assert head.next.val == 3
assert head.next.next.val == 4

head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
head.next.next.next = ListNode(4)
test.deleteNode(head.next)
assert head.val == 1
assert head.next.val == 3
assert head.next.next.val == 4

head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
head.next.next.next = ListNode(4)
test.deleteNode(head.next.next)
assert head.val == 1
assert head.next.val == 2
assert head.next.next.val == 4

head = ListNode(1)
head.next = ListNode(2)
test.deleteNode(head)
assert head.val == 2
print('All Passed!')

Big O Analysis

Space Complexity: O(1)

Time Complexity: O(1)