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merged 1 commit into from
Feb 3, 2024
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feat: add solutions to lcci problem: No.02.01 #2305

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160 changes: 76 additions & 84 deletions lcci/02.01.Remove Duplicate Node/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -34,7 +34,17 @@

## 解法

### 方法一
### 方法一：哈希表

我们创建一个哈希表 $vis$，用于记录已经访问过的节点的值。

然后我们创建一个虚拟节点 $pre$，使得 $pre.next = head$。

接下来我们遍历链表，如果当前节点的值已经在哈希表中，我们就将当前节点删除，即 $pre.next = pre.next.next$；否则，我们将当前节点的值加入哈希表中，并将 $pre$ 指向下一个节点。

遍历结束后，我们返回链表的头节点。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为链表的长度。

<!-- tabs:start -->

Expand All @@ -48,18 +58,14 @@

class Solution:
def removeDuplicateNodes(self, head: ListNode) -> ListNode:
if head is None or head.next is None:
return head
cache = set()
cache.add(head.val)
cur, p = head, head.next
while p:
if p.val not in cache:
cur.next = p
cur = cur.next
cache.add(p.val)
p = p.next
cur.next = None
vis = set()
pre = ListNode(0, head)
while pre.next:
if pre.next.val in vis:
pre.next = pre.next.next
else:
vis.add(pre.next.val)
pre = pre.next
return head
```

Expand All @@ -74,20 +80,15 @@ class Solution:
*/
class Solution {
public ListNode removeDuplicateNodes(ListNode head) {
if (head == null || head.next == null) {
return head;
}
Set<Integer> s = new HashSet<>();
s.add(head.val);
ListNode cur = head;
for (ListNode p = head.next; p != null; p = p.next) {
if (!s.contains(p.val)) {
cur.next = p;
cur = cur.next;
s.add(p.val);
Set<Integer> vis = new HashSet<>();
ListNode pre = new ListNode(0, head);
while (pre.next != null) {
if (vis.add(pre.next.val)) {
pre = pre.next;
} else {
pre.next = pre.next.next;
}
}
cur.next = null;
return head;
}
}
Expand All @@ -105,37 +106,38 @@ class Solution {
class Solution {
public:
ListNode* removeDuplicateNodes(ListNode* head) {
if (head == nullptr || head->next == nullptr) {
return head;
}
unordered_set<int> cache = {head->val};
ListNode* cur = head;
for (ListNode* p = head->next; p != nullptr; p = p->next) {
if (!cache.count(p->val)) {
cur->next = p;
cur = cur->next;
cache.insert(p->val);
unordered_set<int> vis;
ListNode* pre = new ListNode(0, head);
while (pre->next) {
if (vis.count(pre->next->val)) {
pre->next = pre->next->next;
} else {
vis.insert(pre->next->val);
pre = pre->next;
}
}
cur->next = nullptr;
return head;
}
};
```

```go
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func removeDuplicateNodes(head *ListNode) *ListNode {
if head == nil {
return nil
}
vis := map[int]bool{head.Val: true}
p := head
for p.Next != nil {
if vis[p.Next.Val] {
p.Next = p.Next.Next
vis := map[int]bool{}
pre := &ListNode{0, head}
for pre.Next != nil {
if vis[pre.Next.Val] {
pre.Next = pre.Next.Next
} else {
vis[p.Next.Val] = true
p = p.Next
vis[pre.Next.Val] = true
pre = pre.Next
}
}
return head
Expand All @@ -156,17 +158,14 @@ func removeDuplicateNodes(head *ListNode) *ListNode {
*/

function removeDuplicateNodes(head: ListNode | null): ListNode | null {
if (head == null) {
return head;
}
const set = new Set<number>([head.val]);
let cur = head;
while (cur.next != null) {
if (set.has(cur.next.val)) {
cur.next = cur.next.next;
const vis: Set<number> = new Set();
let pre: ListNode = new ListNode(0, head);
while (pre.next) {
if (vis.has(pre.next.val)) {
pre.next = pre.next.next;
} else {
set.add(cur.next.val);
cur = cur.next;
vis.add(pre.next.val);
pre = pre.next;
}
}
return head;
Expand All @@ -193,24 +192,21 @@ function removeDuplicateNodes(head: ListNode | null): ListNode | null {
use std::collections::HashSet;

impl Solution {
pub fn remove_duplicate_nodes(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
match head {
None => head,
Some(mut head) => {
let mut set = HashSet::new();
set.insert(head.val);
let mut pre = &mut head;
while let Some(cur) = &pre.next {
if set.contains(&cur.val) {
pre.next = pre.next.take().unwrap().next;
} else {
set.insert(cur.val);
pre = pre.next.as_mut().unwrap();
}
}
Some(head)
pub fn remove_duplicate_nodes(mut head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
let mut vis = HashSet::new();
let mut pre = ListNode::new(0);
pre.next = head;
let mut cur = &mut pre;
while let Some(node) = cur.next.take() {
if vis.contains(&node.val) {
cur.next = node.next;
} else {
vis.insert(node.val);
cur.next = Some(node);
cur = cur.next.as_mut().unwrap();
}
}
pre.next
}
}
```
Expand All @@ -228,20 +224,16 @@ impl Solution {
* @return {ListNode}
*/
var removeDuplicateNodes = function (head) {
if (head == null || head.next == null) return head;
const cache = new Set([]);
cache.add(head.val);
let cur = head,
fast = head.next;
while (fast !== null) {
if (!cache.has(fast.val)) {
cur.next = fast;
cur = cur.next;
cache.add(fast.val);
const vis = new Set();
let pre = new ListNode(0, head);
while (pre.next) {
if (vis.has(pre.next.val)) {
pre.next = pre.next.next;
} else {
vis.add(pre.next.val);
pre = pre.next;
}
fast = fast.next;
}
cur.next = null;
return head;
};
```
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