思路1

处理头节点，然后遍历下一个节点，只有确保下一个节点不是要移除的节点时再跳到下一个节点

代码1

class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        while(head != nullptr && head->val == val){
            head = head->next;
        }
        ListNode* front = head;

        while(front != nullptr &&front->next != nullptr){
            if(front->next!=nullptr && front->next->val == val){
                front->next = front->next->next;
            }else
                front = front->next;
        }

        return head;
    }
};

思路2

设置空白头节点，这样不用单独处理头节点

代码2

class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode* bankNode = new ListNode();
        bankNode->next = head;
        ListNode* cur = bankNode;

        while(cur->next != nullptr){
            if(cur->next->val == val){
                cur->next = cur->next->next;
            }else
                cur = cur->next;
        }

        return bankNode->next;
    }
};