203. Remove Linked List Elements

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: []

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 {
 *     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) {
        struct ListNode*dummyHead=new ListNode(0,head);
        struct ListNode*pre=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