876. 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.

Rust Solution

struct Solution;
use rustgym_util::*;

struct List {
    head: ListLink,
}

impl List {
    fn new(head: ListLink) -> Self {
        List { head }
    }

    fn middle(&self) -> &ListLink {
        let mut slow = &self.head;
        let mut fast = &self.head;
        while fast.is_some() && fast.as_ref().unwrap().next.is_some() {
            slow = &slow.as_ref().unwrap().next;
            fast = &fast.as_ref().unwrap().next.as_ref().unwrap().next;
        }
        slow
    }
}

impl Solution {
    fn middle_node(head: ListLink) -> ListLink {
        let list = List::new(head);
        let middle: &ListLink = list.middle();
        middle.clone()
    }
}

#[test]
fn test() {
    let head = list![1, 2, 3, 4, 5];
    let middle = list![3, 4, 5];
    assert_eq!(Solution::middle_node(head), middle);
    let head = list![1, 2, 3, 4, 5, 6];
    let middle = list![4, 5, 6];
    assert_eq!(Solution::middle_node(head), middle);
}

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