206. 反转链表

题目描述

给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。

解题思路

这就很简单了，一边出栈 一边入

#![allow(unused)]

fn main() {
struct Solution {}

// Definition for singly-linked list.
#[derive(PartialEq, Eq, Clone, 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(mut head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        let mut dummy = Box::new(ListNode { val: 0, next: None });
        while let Some(mut node) = head {
            let tmp = dummy.next.take();
            let n = node.next.take();
            node.next = tmp;
            dummy.next = Some(node);
            head = n;
        }
        dummy.next
    }
}
}

学习感想

所以我这个算是什么方式

#![allow(unused)]


fn main() {
struct Solution {}

// Definition for singly-linked list.
#[derive(PartialEq, Eq, Clone, 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(mut head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        let mut res: Option<Box<ListNode>> = None;
        while let Some(mut x) = head {
            head = x.next.take();
            x.next = res;
            res = Some(x);
        }
        res
    }
}
}