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. Vickey says:

/**
* 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;
}
}