算法：獲取鏈表的中間值Middle of the Linked List

算法

876. Middle of the Linked List

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.

解法

解決思路：fast快, slow慢兩個鏈表同時走, fast走兩步，slow走一步。當fast爲null時，則slow剛好在中間位置。

比如：[1,2,3,4,5]

第0步： slow 1, fast 1
第1步： slow 2, fast 3
第2步： slow 3, fast 5

代碼實現

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode middleNode(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;
        while (fast != null && fast.next != null) {
          slow = slow.next;
          fast = fast.next.next;
        }
        return slow;
    }
}