notes/CS notes/pages/Leetcode Reverse-Linked-List.md

Leetcode Reverse-Linked-List

2022-06-15 22:07

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Algorithms:

#algorithm #recursion #iteration

Data structures:

#DS #linked_list

Difficulty:

#leetcode #coding_problem #difficulty-medium

Lists:

#CS_list_need_understanding #CS_list_need_practicing

Related topics:

tag:#linked_list

• Floyd's Cycle Finding Algorithm
• Leetcode Linked-List-Cycle
• Leetcode Merge-Two-Sorted-Lists
• Leetcode Remove-Linked-List-Elements
• Two pointers approach

Links:

• Link to problem

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Problem

Given the head of a singly linked list, reverse the list, and return the reversed list.

Examples

Example 1:

**Input:** head = [1,2,3,4,5]
**Output:** [5,4,3,2,1]

Example 2:

**Input:** head = [1,2]
**Output:** [2,1]

Example 3:

**Input:** head = []
**Output:** []

Constraints

• The number of nodes in the list is the range [0, 5000].
• -5000 <= Node.val <= 5000

Thoughts

I thought a slow O(n ^ 2) hybrid solution, while there are better algorithms, using in-place insert, or recursion.

The in place insert is easier to understand, and simple to implement, using a very clever trick.

[!summary] My thoughts on the recursion algorithm

How was that implemented?

The recursion algorithm takes the advantage of the fact that when you change the node's next properties or nodes after that, the node before still points to the same node, so when every node after tmp is reversed, simply move tmp after tmp->next, which now points to the tail of reversed list

Why return the last element of the original list?

It returns the last element in the original list to make sure you are always returning the head of the reversed array, since for any reversed list, the last one comes first.

Solution

I've referred to this guy: https://leetcode.com/problems/reverse-linked-list/discuss/58130/C%2B%2B-Iterative-and-Recursive

This code is hard to understand. ==TODO==: Make my own version

class Solution {
public:
  ListNode *reverseList(ListNode *head) {
    ListNode *pre = new ListNode(0), *cur = head;
    // pre is before head, and insert any element after pre.
    pre->next = head;
    while (cur && cur->next) {
      // temp points to head
      ListNode *temp = pre->next;
      // Move cur->next after pre.
      pre->next = cur->next;
      // Fix pointers, because cur->next is unchanged when changing position.
      cur->next = cur->next->next;
      pre->next->next = temp;
    }
    return pre->next;
  }
};

Recursion: ==TODO== Needs revision!

class Solution {
public:
  ListNode *reverseList(ListNode *head) {
    // Base case: reached the end of list
    if (!head || !(head->next)) {
      return head;
    }

    // node is the end of the initial linked list, which stays the same
    ListNode *node = reverseList(head->next);

    // move head after head->next
    // notice that the node before head still points to head!
    head->next->next = head;

    // since the order is reversed, head->next now points to the tail of the linked list. This is the genius part of this algorithm
    head->next = NULL;

    return node;
  }
};