题解 | #反转链表# | Rust
反转链表
https://www.nowcoder.com/practice/75e878df47f24fdc9dc3e400ec6058ca
/** * #[derive(PartialEq, Eq, Debug, Clone)] * pub struct ListNode { * pub val: i32, * pub next: Option<Box<ListNode>> * } * * impl ListNode { * #[inline] * fn new(val: i32) -> Self { * ListNode { * val: val, * next: None, * } * } * } */ struct Solution{ } impl Solution { fn new() -> Self { Solution{} } /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param head ListNode类 * @return ListNode类 */ pub fn ReverseList(&self, head: Option<Box<ListNode>>) -> Option<Box<ListNode>> { let mut ans = ListNode{val:-1, next:None}; let mut head = head; while head.is_some() { let head_next = head.as_mut().unwrap().next.take(); let ans_next = ans.next.take(); ans.next = head; ans.next.as_mut().unwrap().next = ans_next; head = head_next; } return ans.next; } }