# Posts published in “List”

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

Example 1:

Input: 1->2->3->3->4->4->5
Output: 1->2->5


Example 2:

Input: 1->1->1->2->3
Output: 2->3

## Solution:

Time complexity: O(n)
Space complexity: O(1)

## Related Problems

Given a sorted linked list, delete all duplicates such that each element appear only once.

Example 1:

Input: 1->1->2
Output: 1->2


Example 2:

Input: 1->1->2->3->3
Output: 1->2->3

## Solution:

Time complexity: O(n)
Space complexity: O(1)

## C++

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

Example:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

Time complexity: O(n)
Space complexity: O(1)

## C++

Given the head of a linked list, we repeatedly delete consecutive sequences of nodes that sum to 0until there are no such sequences.

After doing so, return the head of the final linked list.  You may return any such answer.

(Note that in the examples below, all sequences are serializations of ListNode objects.)

Example 1:

Input: head = [1,2,-3,3,1]
Output: [3,1]
Note: The answer [1,2,1] would also be accepted.


Example 2:

Input: head = [1,2,3,-3,4]
Output: [1,2,4]


Example 3:

Input: head = [1,2,3,-3,-2]
Output: 


Constraints:

• The given linked list will contain between 1 and 1000 nodes.
• Each node in the linked list has -1000 <= node.val <= 1000.

## Solution: HashTable

Similar to target sum = 0, use a hashtable to store the first ListNode* that has a given prefix sum. Whenever the same prefix sum occurs, skip all the elements between the first occurrence and current one, e.g. first_sum_x.next = curr_sum_x.next

Time complexity: O(n)
Space complexity: O(n)

## Python3

Reverse a linked list from position m to n. Do it in one-pass.

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

Solution:

1. Store the m – 1 and m-th item as prev and tail before reversing
2. Reverse the m to n, return the head and tail->next of the reversed list
3. Reconnect prev and head, tail and tail->next

Time complexity: O(n)
Space complexity: O(1)

## Python3

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