0800 - 0899
0876 - Middle of the Linked List (Easy)
Problem Link
https://leetcode.com/problems/middle-of-the-linked-list/description/
Problem Statement
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.
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
Approach 1: Fast and Slow Pointer
Classic Fast and Slow Pointer question
- Time Complexity: where is the number of nodes
- Space Complexity:
/**
* 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) {
// fast & slow pointer
// slow moves 1 step
// fast moves 2 steps
// 1 -> 2 -> 3 -> 4 -> 5
// slow : 1 -> 2 -> 3
// fast : 1 -> 3 -> 5
// when fast reaches the end, slow will be the middle
ListNode* slow = head;
ListNode* fast = head;
while (fast != NULL && fast->next != NULL) {
slow = slow->next;
fast = fast->next->next;
}
return slow;
}
};