Leetcode 237.Delete Node in a Linked List
===
### Description
(easy)

Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and return the new head.

#### example 1:

Input: head = [1,2,6,3,4,5,6], val = 6
Output: [1,2,3,4,5]
#### example 2:
Input: head = [], val = 1
Output: []
#### Example 3:
Input: head = [7,7,7,7], val = 7
Output: []

### 想法
c在當整個list清為[]的時候，需要多一個判斷式把整個清掉。
python多設一個dummy節點在head之前，然後while下去比較，若是cur.next.val = val的話，cur.next = cur.next.next，用這個方法就可以少一些判斷式，比如說當head=[]的時候。但是這個方法為多加一個節點，多用一些空間

### 程式碼

#### c:

struct ListNode* removeElements(struct ListNode* head, int val){
    struct ListNode* current=head;
  if(head==NULL) return head;
  while(current!=NULL){
    if(current->next==NULL) break;
    if(current->next->val==val){
      current->next=current->next->next;
    }else{
      current=current->next;
    }
  }
  if(head->val==val){
    current=head;
    head=head->next;
    free(current);
  }
  return head;
}

Runtime: 17 ms
Memory Usage: 8.1 MB

#### python:

class Solution:
    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
        dummy = cur = ListNode(-1)
        cur.next = head     
        while cur.next:
            if cur.next.val == val:
                cur.next = cur.next.next
            else:
                cur = cur.next
            head = head.next
        
        return dummy.next

Runtime: 89 ms
Memory Usage: 17.9 MB