83 lines
1.8 KiB
Markdown
83 lines
1.8 KiB
Markdown
# Leetcode Reverse-Linked-List
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#### 2022-06-15 22:07
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---
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##### Algorithms:
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#algorithm #recursion #iteration
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##### Data structures:
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#DS #linked_list
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##### Difficulty:
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#leetcode #coding_problem #difficulty-easy
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##### Lists:
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#CS_list_need_understanding #CS_list_need_practicing
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##### Related topics:
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```expander
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tag:#linked_list
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```
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##### Links:
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- [Link to problem](https://leetcode.com/problems/reverse-linked-list/)
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___
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### Problem
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Given the `head` of a singly linked list, reverse the list, and return _the reversed list_.
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#### Examples
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**Example 1:**
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```markdown
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**Input:** head = [1,2,3,4,5]
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**Output:** [5,4,3,2,1]
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```
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**Example 2:**
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```markdown
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**Input:** head = [1,2]
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**Output:** [2,1]
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```
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**Example 3:**
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```markdown
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**Input:** head = []
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**Output:** []
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```
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#### Constraints
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- The number of nodes in the list is the range `[0, 5000]`.
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- `-5000 <= Node.val <= 5000`
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### Thoughts
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I thought a slow O(n ^ 2) hybrid solution, while there are better algorithms, using in-place insert, or recursion.
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The in place insert is easier to understand, and simple to implement, using a very clever trick.
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### Solution
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I've referred to this guy: https://leetcode.com/problems/reverse-linked-list/discuss/58130/C%2B%2B-Iterative-and-Recursive
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```cpp
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class Solution {
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public:
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ListNode *reverseList(ListNode *head) {
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ListNode *pre = new ListNode(0), *cur = head;
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// pre is before head, and insert any element after pre.
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pre->next = head;
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while (cur && cur->next) {
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// Move cur->next after pre.
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ListNode *temp = pre->next;
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pre->next = cur->next;
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cur->next = cur->next->next;
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pre->next->next = temp;
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}
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return pre->next;
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}
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};
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``` |