203. 移除链表元素

题目

很长时间没接触链表这种数据结构了，看到这道题有点无从下手。

看了卡哥讲解视频

特别想说的是，卡哥的视频把这个题讲的很好，有些不理解的地方，怎么思考的，卡哥讲的很清楚。

跟着卡哥代码敲了一下：

方法一：不设置虚拟头节点

/**
 * 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) {
        while (head != nullptr && head->val == val)
        {
            ListNode *tmp = head;
            head = head->next;
            delete tmp;
        }

        ListNode *cur = head;
        while (cur != nullptr && cur->next != nullptr)
        {
            if (cur->next->val == val)
            {
                ListNode *tmp = cur->next;
                cur->next = cur->next->next;
                delete tmp;
            }
            else 
            {
                cur = cur->next;
            }
        }
        return head;
    }
};

附上一张用于理解的图：

方法二：设置虚拟头节点

/**
 * 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 *dummyHead = new ListNode(0);
        dummyHead->next = head;
        ListNode *cur = dummyHead;
        while (cur->next != nullptr)
        {
            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;
    }
};