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.
代码
Approach #1 Output to Array
Time O(N) Space O(N)
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */classSolution {publicListNodemiddleNode(ListNode head) {ListNode[] A =newListNode[100];int t =0;while (head !=null) {A[t++] = head; head =head.next; }returnA[t /2]; }}
Approach #2 Fast and Slow Pointer
Time O(N) Space O(1)
classSolution {publicListNodemiddleNode(ListNode head) {ListNode slow = head, fast = head;while (fast !=null&&fast.next!=null) { slow =slow.next; fast =fast.next.next; }return slow; }}