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 and 100.
Solution
Idea is to use slow and fast (turtle and rabbit) two nodes from head, slow node move 1 node each time,
and fast node move 2 nodes each time, until fast is out of node, slow node is in the middle of the list node, return slow.
Time complexity:
O(N)
N - the length of linkedlist
//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;
ListNode fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
}
