LeetCode 876. Middle of the Linked List

Description

https://leetcode.com/problems/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.

Note:

  • The number of nodes in the given list will be between 1 and 100.

Explanation

fast pointer to find the middle number. slow pointer to get to the middle node.

Python Solution

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def middleNode(self, head: ListNode) -> ListNode:
        if head == None:
            return head
        
        count = 0
        
        fast = head
        while fast != None:
            fast = fast.next
            count += 1
        
        mid = count // 2 + 1
        
        slow = head
        for i in range(1, mid):
            slow = slow.next
            
        return slow
  • Time complexity: O(n).
  • Space complexity: O(1).

3 Thoughts to “LeetCode 876. Middle of the Linked List”

  1. /**
    * 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 end = head;
    ListNode mid = head;
    while(end!=null && end.next!=null){
    end = end.next.next;
    mid = mid.next;
    }
    return mid;
    }

    public ListNode middleNode1(ListNode head) {

    ListNode midNode = head;
    int len = getLength(head);
    int mid = len/2;

    while(mid > 0){
    midNode = head.next;
    head = head.next;
    mid–;
    }

    return midNode;
    }

    public int getLength(ListNode head){
    int count =0;
    while(head != null){
    head = head.next;
    count++;
    }
    return count;
    }
    }

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