head of a singly linked list, return the middle node of the linked list.
If there are two middle nodes, return the second middle node.
Input: head = [1,2,3,4,5] Output: [3,4,5] Explanation: The middle node of the list is node 3.
Input: head = [1,2,3,4,5,6] Output: [4,5,6] Explanation: Since the list has two middle nodes with values 3 and 4, we return the second one.
- The number of nodes in the list is in the range
1 <= Node.val <= 100
# Definition for singly-linked list. # class ListNode(object): # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution(object): def middleNode(self, head): """ :type head: ListNode :rtype: ListNode """ """ We can do this with 2 loops, can we do it with just 1 loop? I am thinking 2 pointers First pointer is "fast", it travels at N speed Second pointer travels at N/2 speed So it only increments when pointer1 is > n/2 away Because by definition the "middle" is half of the max length therefore the maxium the second pointer can travel is n / 2 Alternatively we can view this a different way. The maximum speed pointer1 can travel is N the maximum speed pointer2 can travel is half of N, so n / 2 Therefore we use a fast and slow approach """ fast, slow = head, head while fast and fast.next: # slow will happen to be in the middle when fast can no longer run fast = fast.next.next slow = slow.next return slow
This is similar to the fast and slow problem: