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 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93
//! Reverse Linked List [leetcode: reverse_linked_list](https://leetcode.com/problems/reverse-linked-list) //! //! Reverse a singly linked list. //! //! ***Example:*** //! //! ``` //! Input: 1->2->3->4->5->NULL //! //! Output: 5->4->3->2->1->NULL //! ``` //! //! ***Follow up:*** //! //! A linked list can be reversed either iteratively or recursively. Could you implement both? /// # Solutions /// /// # Approach 1: Iterative /// /// * Time complexity: O(n) /// /// * Space complexity: O(1) /// /// ```rust /// /// // Definition for singly-linked list. /// // #[derive(PartialEq, Eq, Debug)] /// // pub struct ListNode { /// // pub val: i32, /// // pub next: Option<Box<ListNode>> /// // } /// // /// // impl ListNode { /// // #[inline] /// // fn new(val: i32) -> Self { /// // ListNode { /// // next: None, /// // val /// // } /// // } /// // } /// /// impl Solution { /// pub fn reverse_list(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> { /// if head.is_none() { return None; } /// /// let mut prev = None; /// let mut current = head; /// while let Some(mut tmp) = current.take() { /// let next = tmp.next.take(); /// tmp.next = prev.take(); /// prev = Some(tmp); /// current = next; /// } /// /// prev /// } /// } /// ``` /// pub fn reverse_list(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> { if head.is_none() { return None; } let mut prev = None; let mut current = head; while let Some(mut tmp) = current.take() { let next = tmp.next.take(); tmp.next = prev.take(); prev = Some(tmp); current = next; } prev } // Definition for singly-linked list. #[derive(PartialEq, Eq, Debug)] pub struct ListNode { pub val: i32, pub next: Option<Box<ListNode>> } #[allow(dead_code)] impl ListNode { #[inline] fn new(val: i32) -> Self { ListNode { next: None, val } } }