notes/OJ 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

Revisions:

2022-07-02

Related topics:

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. The only recursive part is returning the tail node.

Solution

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

Insertion, iteration

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:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
  ListNode *reverseList(ListNode *head) {
    // Recursion
    if (head == nullptr || head->next == nullptr) {
      // base case: return the tail
      return head;
    }

    // have to call itself before modifying, since head->next will point to
    // nullptr
    ListNode *newHead = reverseList(head->next);

    head->next->next = head;
    head->next = nullptr;

    return newHead;
  }
};