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:
Input으로 Node가 주어질 때 가운데 있는 노드를 반환하는 문제이다.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
ListNode 클래스는 위와 같이 구현되어있다.
class Solution {
public ListNode middleNode(ListNode head) {
ListNode node = head;
ListNode resultNode = head;
while (resultNode != null && resultNode.next != null) {
node = node.next;
resultNode = resultNode.next.next;
}
return node;
}
}