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:
1 and 100.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);
}