LeetCode 876. Middle of the Linked List

原题链接在这里：https://leetcode.com/problems/middle-of-the-linked-list/

题目：

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.

题解：

walker每次跳一步. runner每次跳两步. runner 跳到尾时, walker就是中点.

list偶数长度时, 中点找中间两个前者时, while的终止条件到是(runner != null && runner.next != null && runner.next.next != null).

找后者时, while的终止条件是(runner != null && runner.next != null).

Time Complexity: O(n).

Space: O(1).

AC Java:

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 class Solution {
10     public ListNode middleNode(ListNode head) {
11         if(head == null || head.next == null){
12             return head;
13         }
14         
15         ListNode walker = head;
16         ListNode runner = head;
17         while(runner != null && runner.next != null){
18             walker = walker.next;
19             runner = runner.next.next;
20         }
21         
22         return walker;
23     }
24 }

LeetCode 876. Middle of the Linked List

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