每日一题04

每日一题day4

每日一题

链表定义

cpp

struct ListNode{
    int val;
    ListNode *next;
    ListNode(int x) : val(x),next(NULL) {}
}

python

class ListNode:
    def __init__(self, val, next = None):
        self.val = val
        self.next = next

移除链表元素

https://leetcode.cn/problems/remove-linked-list-elements/

cpp

/**
 * 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* removeElements(ListNode* head, int val) {
        while(head != NULL && head->val == val){
            ListNode* temp = head;
            head = head-> next;
            delete temp;
        }
        ListNode* cur = head;
        while(cur != NULL && cur->next != NULL){
            if(cur->next->val == val){
                ListNode* temp = cur->next;
                cur->next = cur->next->next;
                delete temp;
            }else{
                cur = cur->next;
            }
        }
        return head;
    }
};

python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
        while head != None and head.val == val:
            head = head.next
        cur = head
        while cur != None and cur.next != None:
            if cur.next.val == val:
                cur.next = cur.next.next
            else:
                cur = cur.next
        return head

leetcode 707 设计链表

https://leetcode.cn/problems/design-linked-list/

代码：

cpp

```

**python**

```python