203. Remove Linked List Elements

Aurora20052024-06-24 10:43

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:

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Input: head = [1,2,6,3,4,5,6], val = 6
Output: [1,2,3,4,5]

Example 2:
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Input: head = [], val = 1
Output: []

Example 3:
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Input: head = [7,7,7,7], val = 7
Output: []

Constraints:

The number of nodes in the list is in the range [0, 104].
1 <= Node.val <= 50
0 <= val <= 50

/**
- Definition for singly-linked list.
- struct ListNode {
- 复制代码
```
int val;
```
- 复制代码
```
ListNode *next;
```
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```
ListNode() : val(0), next(nullptr) {}
```
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```
ListNode(int x) : val(x), next(nullptr) {}
```
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```
ListNode(int x, ListNode *next) : val(x), next(next) {}
```
- };
  /
  class Solution {
  public:
  ListNode removeElements(ListNode* head, int val) {
  struct ListNodedummyHead=new ListNode(0,head);
  struct ListNodepre=dummyHead;
  while(pre->next!=NULL){
  if(pre->next->val==val){
  pre->next=pre->next->next;
  }else{
  pre=pre->next;
  }
  }
  return dummyHead->next;
  }
  };

注意：

1.其实这道题可以有两种方法去做，第一种是不适用虚拟头节点的，但是这种方法需要分类，一种是删除的节点是头节点的时候，第二种是其他元素，但是这样的话，代码不够简洁。所以采用了虚拟头节点的方式来做

2.C++中应该在用完dummyHead后，将其释放，但是这里我忘了