上次编辑到这里，代码来自缓存 点击恢复默认模板
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* modifiedList(vector<int>& nums, ListNode* head) {
}
};
php 解法, 执行用时: 322 ms, 内存消耗: 57.8 MB, 提交时间: 2024-07-15 09:53:45
/**
* Definition for a singly-linked list.
* class ListNode {
* public $val = 0;
* public $next = null;
* function __construct($val = 0, $next = null) {
* $this->val = $val;
* $this->next = $next;
* }
* }
*/
class Solution {
/**
* @param Integer[] $nums
* @param ListNode $head
* @return ListNode
*/
function modifiedList($nums, $head) {
$set = array_flip($nums);
$dummy = new ListNode(0, $head);
$cur = $dummy;
while ( $cur->next !== null ) {
if ( isset($set[$cur->next->val]) ) {
$cur->next = $cur->next->next;
} else {
$cur = $cur->next;
}
}
return $dummy->next;
}
}
javascript 解法, 执行用时: 422 ms, 内存消耗: 88.6 MB, 提交时间: 2024-07-15 09:45:04
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {number[]} nums
* @param {ListNode} head
* @return {ListNode}
*/
var modifiedList = function(nums, head) {
// 创建一个哈希集合来存储nums中的元素
const has = new Set(nums);
// 创建一个哑节点（dummy node）来简化头节点的处理
const dummy = new ListNode(0, head);
let cur = dummy;
// 遍历链表
while (cur.next !== null) {
// 如果当前节点的值在nums数组中，则删除该节点
if (has.has(cur.next.val)) {
cur.next = cur.next.next;
} else {
// 否则，移动到下一个节点
cur = cur.next;
}
}
// 返回修改后的链表的头节点（跳过哑节点）
return dummy.next;
};
golang 解法, 执行用时: 252 ms, 内存消耗: 11.3 MB, 提交时间: 2024-07-15 09:42:26
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func modifiedList(nums []int, head *ListNode) *ListNode {
has := make(map[int]bool, len(nums)) // 预分配空间
for _, x := range nums {
has[x] = true
}
dummy := &ListNode{Next: head}
cur := dummy
for cur.Next != nil {
if has[cur.Next.Val] {
cur.Next = cur.Next.Next // 删除
} else {
cur = cur.Next // 向后移动
}
}
return dummy.Next
}
java 解法, 执行用时: 19 ms, 内存消耗: 65.1 MB, 提交时间: 2024-07-15 09:42:06
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode modifiedList(int[] nums, ListNode head) {
Set<Integer> set = new HashSet<>(nums.length); // 预分配空间
for (int x : nums) {
set.add(x);
}
ListNode dummy = new ListNode(0, head);
ListNode cur = dummy;
while (cur.next != null) {
if (set.contains(cur.next.val)) {
cur.next = cur.next.next; // 删除
} else {
cur = cur.next; // 向后移动
}
}
return dummy.next;
}
}
cpp 解法, 执行用时: 474 ms, 内存消耗: 256.7 MB, 提交时间: 2024-07-15 09:41:54
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* modifiedList(vector<int>& nums, ListNode* head) {
unordered_set<int> st(nums.begin(), nums.end());
ListNode dummy(0, head);
ListNode* cur = &dummy;
while (cur->next) {
auto nxt = cur->next;
if (st.contains(nxt->val)) {
cur->next = nxt->next; // 删除
delete nxt; // 释放内存
} else {
cur = nxt; // 向后移动
}
}
return dummy.next;
}
};
cpp 解法, 执行用时: 445 ms, 内存消耗: 252.5 MB, 提交时间: 2024-07-15 09:41:45
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* modifiedList(vector<int>& nums, ListNode* head) {
unordered_set<int> st(nums.begin(), nums.end());
ListNode dummy(0, head);
ListNode* cur = &dummy;
while (cur->next) {
if (st.contains(cur->next->val)) {
cur->next = cur->next->next; // 删除
} else {
cur = cur->next; // 向后移动
}
}
return dummy.next;
}
};
python3 解法, 执行用时: 364 ms, 内存消耗: 56.4 MB, 提交时间: 2024-07-15 09:41:28
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def modifiedList(self, nums: List[int], head: Optional[ListNode]) -> Optional[ListNode]:
st = set(nums)
cur = dummy = ListNode(next=head)
while cur.next:
if cur.next.val in st:
cur.next = cur.next.next # 删除
else:
cur = cur.next # 向后移动
return dummy.next
