一、题目

二、思路

基础题，从头到尾遍历链表节点元素，遇到目标节点就删除。

一开始没用哨兵节点，处理第一个元素是目标结点，和处理第一个和第二个元素为目标结点，这两种情况处理起来有点冗余，，只过了44个测试用例：

/**
 * 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* slow = head;
        ListNode* fast = head;

        if(fast == NULL){
            return head;
        }
        fast = fast->next;
        if(fast == NULL && slow->val == val){
            return NULL;
        }else if(fast != NULL && slow->val == val){
            slow = slow ->next;
            fast = fast->next;
            head = slow->next;
        }
        while(fast != NULL){
            if(fast->val == val){
                slow->next = fast->next;
                fast = fast->next;
                continue;
            }
            fast = fast->next;
            slow = slow->next;
        }
        return head;
    }
};

后面想起要介绍哨兵节点后就舒服了：

/**
 * 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* fake = new ListNode(0);
        fake->next = head;
        ListNode* slow = fake;
        ListNode* fast = fake->next;
        
        if(fast == NULL){
            return head;
        }
        while(fast != NULL){
            if(fast->val == val){
                slow->next = fast->next;
                fast = fast->next;
                continue;
            }
            fast = fast->next;
            slow = slow->next;
        }
        return fake->next;
    }
};