lc430
DFS遍历链表
把每个节点的++子链表"拉直"插入到当前节点和下一个节点之间++
最终返回扁平化后的头节点
class Solution {
public:
Node* dfs(Node *now){
if(now == NULL) return NULL;
Node* ans = NULL;
if(now->child){
Node *t1 = now->child;
Node *t2 = dfs(now->child);
++Node *t3 = now->next;++
now->child = NULL;
now->next = t1;
t1->prev = now;
++t2->next = t3;++
if(t3 == NULL) return t2;
t3->prev = t2;
++ans = dfs(t3);++
}
else
{
if(now->next == NULL) return now;
ans = dfs(now->next);
}
return ans;
}
Node* flatten(Node* head) {
dfs(head);
return head;
}
};
rust
#derive(Debug, Clone, PartialEq, Eq)
pub struct Node {
pub val: i32,
pub prev: Option<Rc<RefCell<Node>>>,
pub next: Option<Rc<RefCell<Node>>>,
pub child: Option<Rc<RefCell<Node>>>,
}
impl Node {
#inline
pub fn new(val: i32) -> Self {
Node {
val,
prev: None,
next: None,
child: None,
}
}
}
use std::rc::Rc;
use std::cell::RefCell;
struct Solution;
impl Solution {
pub fn flatten(head: Option<Rc<RefCell<Node>>>) -> Option<Rc<RefCell<Node>>> {
if let Some(head) = head.clone() {
Self::dfs(head);
}
head
}
fn dfs(node: Rc<RefCell<Node>>) -> Rc<RefCell<Node>> {
let mut node_mut = node.borrow_mut();
if let Some(child) = node_mut.child.take() {
let next = node_mut.next.clone();
let child_tail = Self::dfs(child.clone());
node_mut.next = Some(child.clone());
child.borrow_mut().prev = Some(node.clone());
child_tail.borrow_mut().next = next.clone();
if let Some(next) = next {
next.borrow_mut().prev = Some(child_tail.clone());
}
if let Some(next) = next {
drop(node_mut);
return Self::dfs(next);
} else {
return child_tail;
}
}
if let Some(next) = node_mut.next.clone() {
drop(node_mut);
Self::dfs(next)
} else {
drop(node_mut);
node
}
}
}
-
Rc<RefCell>：共享所有权+内部可变性，安全管理链表节点的引用与修改；
-
Option枚举：优雅处理空指针场景，避免野指针；
-
take()方法：转移child字段所有权，确保扁平化后原节点无残留子节点；
-
borrow_mut()/drop()：严格遵循借用规则，避免可变借用冲突；
-
模式匹配：通过if let简化Option类型的解构与判断。
lcr154
哈希表存原节点到新节点的映射
先复制节点值，再补全next和random指针
最后返回新链表头--hash val
class Solution {
public:
Node* copyRandomList(Node* head) {
if(head == nullptr) return nullptr;
Node* cur = head;
unordered_map<Node*, Node*> map;
// 3. 复制各节点，并建立 "原节点 -> 新节点" 的 Map 映射
while(cur != nullptr) {
mapcur = new Node(cur->val);
cur = cur->next;
}
cur = head;
// 4. 构建新链表的 next 和 random 指向
while(cur != nullptr) {
mapcur->next = mapcur-\>next;
mapcur->random = mapcur-\>random;
cur = cur->next;
}
// 5. 返回新链表的头节点
return maphead;
}
};
rust
把原链表节点和复制节点穿插排列，先补复制节点的random指针，再拆分出复制链表（不占额外hash空间）
// Definition for a Node.
// #derive(Debug, Clone, PartialEq, Eq)
// pub struct Node {
// pub val: i32,
// pub next: Option<Rc<RefCell<Node>>>,
// pub random: Option<Rc<RefCell<Node>>>,
// }
//
// impl Node {
// #inline
// pub fn new(val: i32) -> Self {
// Node {
// val,
// next: None,
// random: None,
// }
// }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
pub fn copy_random_list(head: Option<Rc<RefCell<Node>>>) -> Option<Rc<RefCell<Node>>> {
let mut current = head.clone();
// Step 1: Interleave copied nodes with original nodes
while let Some(node) = current {
let copied = Rc::new(RefCell::new(Node::new(node.borrow().val)));
copied.borrow_mut().next = node.borrow().next.clone();
node.borrow_mut().next = Some(copied.clone());
current = copied.borrow().next.clone();
}
// Step 2: Assign random pointers for the copied nodes
current = head.clone();
while let Some(node) = current {
if let Some(random) = node.borrow().random.clone() {
node.borrow().next.clone()?.borrow_mut().random = random.borrow().next.clone();
}
let next = node.borrow().next.clone();
current = next?.borrow().next.clone();
}
// Step 3: Separate the copied list from the original list
current = head.clone();
let new_head = head.clone()?.borrow().next.clone();
while let Some(node) = current {
let next = node.borrow().next.clone();
node.borrow_mut().next = next.clone()?.borrow().next.clone();
if let Some(node_next) = node.borrow().next.clone() {
next?.borrow_mut().next = node_next.borrow().next.clone();
}
current = node.borrow().next.clone();
}
new_head
}
}