当前位置: 代码迷 >> 综合 >> LeetCode-Java-876. Middle of the Linked List
  详细解决方案

LeetCode-Java-876. Middle of the Linked List

热度:84   发布时间:2023-12-16 09:11:25.0

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.

代码

先找到位置,然后遍历查找即可

/*** Definition for singly-linked list.* public class ListNode {* int val;* ListNode next;* ListNode(int x) { val = x; }* }*/
class Solution {public ListNode middleNode(ListNode head) {ListNode cur = head;int i = 0;while(cur!=null){i++;cur = cur.next;}for(int j=0;j<i/2;j++){head = head.next;}return head;}
}

第二种方式

这种方式是两个指针一个一次走两步一个一次走一步,当一次走两步的快指针到最后一步的时候,慢指针刚好指向中间位置

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