一道基础的链表相关题目，在删除时对头节点进行单独处理。

/**
 * 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 * t = head;
            head = head->next;
            delete t;
        }
        ListNode * cur = head;
        while(cur != NULL && cur->next != NULL){
            if(cur->next->val == val){
                ListNode * t = cur->next;
                cur->next = cur->next->next;
                delete t;
            }else{
                cur = cur->next;
            }
        }

        return head;
    }
};

另外一种写法是设置一个虚拟节点指向头节点，这样就无需对头节点进行单独处理，最后将head指向虚拟节点的下一个节点。

class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode* dummyHead = new ListNode(0); // 设置一个虚拟头结点
        dummyHead->next = head; // 将虚拟头结点指向head，这样方便后面做删除操作
        ListNode* cur = dummyHead;
        while (cur->next != NULL) {
            if(cur->next->val == val) {
                ListNode* tmp = cur->next;
                cur->next = cur->next->next;
                delete tmp;
            } else {
                cur = cur->next;
            }
        }
        head = dummyHead->next;
        delete dummyHead;
        return head;
    }
};