c++解法

解法1：先确定头节点，而后移动指针

/**
 * 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 != nullptr && head->val == val) head = head->next;
        if(head == nullptr) return nullptr;

        ListNode* pre = head;
        while(pre->next != nullptr){
            if(pre->next->val == val) pre->next = pre->next->next;
            else pre = pre->next;
        }
        return head;
    }
};

解法2：虚拟头节点

/**
 * 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) {
       ListNode* dummy = new ListNode(-1);//虚拟头节点，val值随便赋
       dummy ->next = head;
       ListNode *pre = dummy;
       while(pre->next != nullptr){
           if(pre->next->val == val) pre->next = pre->next->next;
           else pre = pre->next;
       }
       return dummy->next;
    }
};