# Posts tagged as “list”

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

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Example:

Input:
[
1->4->5,
1->3->4,
2->6
]
Output: 1->1->2->3->4->4->5->6

## Solution 1: Min heap

Time complexity: O(nklogk)
Space complexity: O(k)

## Solution 2: Merge Sort

Time complexity: O(nklogk)
Space complexity: O(logk)

# Problem

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

• Only constant extra memory is allowed.
• You may not alter the values in the list’s nodes, only nodes itself may be changed.

# Solution

Two passes.

First pass, get the length of the list.

Second pass, swap in groups.

Time complexity: O(n)

Space complexity: O(1)

# Problem

Example:

Given 1->2->3->4, you should return the list as 2->1->4->3.

Note:

• Your algorithm should use only constant extra space.
• You may not modify the values in the list’s nodes, only nodes itself may be changed.

# Solution

Time complexity: O(n)

Space complexity: O(1)

# Related Problems

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