• 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 }
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  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/10961792.html
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