LeetCode-Java-876. 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.

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;

    }
}