876. Middle of Linked List.java

E
tags: Linked List

找Linked List的中间node

#### 快慢指针
- 不在乎slow是不是到底，因为fast肯定先到。
- 确保fast, fast.next不是Null就好

```

/**
LeetCode
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.
*/


/*
LintCode
Find the middle node of a linked list.

Have you met this question in a real interview? Yes
Example
Given 1->2->3, return the node with value 2.

Given 1->2, return the node with value 1.

Tags Expand 
Linked List

*/
/*
    Thoughts:
    Practice LinkedList, fast and slow pointer
*/

public class Solution {
    /**
     * @param head: the head of linked list.
     * @return: a middle node of the linked list
     */
    public ListNode middleNode(ListNode head) { 
        if (head == null || head.next == null) {
            return head;
        }
        
        ListNode slow = head;
        ListNode fast = head.next;
        while (fast!= null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
}


/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
```