LeetCode #876 Middle of the Linked List

Posted on | Post modified | In , , , , , , , , , ,

Problem

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.

Examples

1
2
3
4
5
6
7
8
9
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.

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.

Solution

Method: Two Pointers

Time Complexity: O(n)

Space Complexity: O(1)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def print(self, head):
        while head:
            print(head.val, end=' ')
            head = head.next

    def middleNode(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        jump_one = jump_two = head
        
        while jump_one and jump_two:
            jump_two = jump_two.next
            if jump_two:
                jump_two = jump_two.next
                jump_one = jump_one.next
                
        return jump_one

# test case
# head = ListNode(1)
# head.next = ListNode(2)
# head.next.next = ListNode(3)
# head.next.next.next = ListNode(4)
# Solution().print(Solution().middleNode(head))