Formatted question description: https://leetcode.ca/all/876.html
876. Middle of the Linked List (Easy)
Given a non-empty, singly linked list with head node head
, return a middle node of linked list.
If there are two middle nodes, return the second middle node.
Example 1:
Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3. (The judge’s serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans, such that:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.
Example 2:
Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization: [4,5,6])
Since the list has two middle nodes with values 3 and 4, we return the second one.
Note:
- The number of nodes in the given list will be between
1
and100
.
Solution 1. Two Pointers
// OJ: https://leetcode.com/problems/middle-of-the-linked-list/
// Time: O(N)
// Space: O(1)
class Solution {
public:
ListNode* middleNode(ListNode* head) {
auto slow = head, fast = head;
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
}
return slow;
}
};
Java
-
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode middleNode(ListNode head) { if (head == null || head.next == null) return head; ListNode slow = head, fast = head; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; } return slow; } }
-
// OJ: https://leetcode.com/problems/middle-of-the-linked-list/ // Time: O(N) // Space: O(1) class Solution { public: ListNode* middleNode(ListNode* head) { auto slow = head, fast = head; while (fast && fast->next) { slow = slow->next; fast = fast->next->next; } return slow; } };
-
# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def middleNode(self, head): """ :type head: ListNode :rtype: ListNode """ dummy = ListNode(0) dummy.next = head slow, fast = dummy, dummy while fast and fast.next: slow = slow.next fast = fast.next.next return slow.next if fast else slow