1.9 KiB
1.9 KiB
Leetcode Middle-of-the-Linked-List
2022-07-13 09:08
Algorithms:
#algorithm #two_pointers
Data structures:
#DS #linked_list
Difficulty:
#coding_problem #difficulty-easy
Additional tags:
#leetcode
Revisions:
N/A
Related topics:
Links:
Problem
Given the head of a singly linked list, return the middle node of the linked list.
If there are two middle nodes, return the second middle node.
Examples
Example 1:
Input: head = [1,2,3,4,5] Output: [3,4,5] Explanation: The middle node of the list is node 3.
Example 2:
Input: head = [1,2,3,4,5,6] Output: [4,5,6] Explanation: Since the list has two middle nodes with values 3 and 4, we return the second one.
Constraints
- The number of nodes in the list is in the range
[1, 100]
. 1 <= Node.val <= 100
Thoughts
[!summary] This is a #two_pointers problem To find the middle of linked list, use two pointers: one fast pointer and one small pointer
Solution
/**
* 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 *middleNode(ListNode *head) {
// two pointers problem.
ListNode *fast, *slow;
fast = slow = head;
while (true) {
if (fast->next && fast->next->next) {
slow = slow->next;
fast = fast->next->next;
} else if (!fast->next) {
return slow;
} else if (!fast->next->next) {
return slow->next;
}
}
}
};