刷题记录Day3（补）

203. 移除链表元素

① 头节点的判断

② 指针移动的判断，判断他的 next，防止非法访问

/**
 * 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;
    }
};

206. 反转链表

① 头节点是否为空的判断

② 指针移动条件

③ 怎么保证指针方向移动后，已改变方向链表和链未改变方向链表的联系

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(head==nullptr)
        {
            return nullptr;
        }
        else
        {
            ListNode* cur = head;
            ListNode* tmp = cur->next;
            cur->next = nullptr;
            while(tmp!=nullptr)
            {
                cur = tmp;
                tmp = tmp->next;
                cur->next = head;
                head = cur;
            }
            return head;
        }
    }
};