[3]-代码随想录算法训练营-day3-链表-part1

代码随想录算法训练营第二天|链表 part1

1.LeetCode 203.移除链表元素

1. 题目

   • https://leetcode.cn/problems/remove-linked-list-elements/

2. 思路

   • 遍历，删
   • 两种处理方式，用虚拟头结点、仅用现有连接

3. 刷随想录后想法

   • 思路开拓了
   • 代码编写简洁

4. 实现困难

   • 头结点处理
   • 连续节点处理超时

5. 实现代码

   /**
    * 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 *cur, *next = nullptr;
           if(!head){
               return head;
           }
   
           //头节点为val
           while(head && val == head->val){
               cur = head;
               head = head->next;
               delete(cur);
           }
   
           //头结点非val
           cur = head;
           while(cur){
               next = cur->next;
               if(next && val == next->val){
                   cur->next = next->next;
                   delete(next);
               }else{
                   cur = cur->next;
               }
           }
           return head;
       }
   };
   

6. 学习收获

   • 处理非头结点时，用else判断cur是否移动
   • C++释放内存用delete，与之对应的开辟空间用new
   • LeetCode错误判断及处理

2.LeetCode 707.设计链表

1. 题目

   • https://leetcode.cn/problems/design-linked-list/

2. 思路

   • 常规思路，使用带头结点的链表

3. 刷随想录后想法

   • 节点处理清晰

4. 实现困难

   • C++语法问题

5. 实现代码

   class MyLinkedList {
   public:
       struct LinkedNode{
           int val;
           LinkedNode *next;
           LinkedNode(int val):val(val), next(nullptr){}
       };
   
       MyLinkedList() {
           //带头结点的链表
           _dummyHead = new LinkedNode(0);
       }
   
       int get(int index) {
   
           int count = 0;
           LinkedNode *cur = _dummyHead->next;
           while(cur){
               if(count == index){
                   return cur->val;
               }else{
                   count++;
                   cur = cur->next;
               }
           }
           return -1;
       }
   
       void addAtHead(int val) {
           if(_dummyHead){
               LinkedNode *node = new LinkedNode(val);
               node->val = val;
               node->next = _dummyHead->next;
               _dummyHead->next = node;
           }
       }
   
       void addAtTail(int val) {
           LinkedNode *node = new LinkedNode(val);
           node->val = val;
           node->next = nullptr;
   
           LinkedNode *cur = _dummyHead;
           while(cur && cur->next){
               cur = cur->next;
           }
           cur->next = node;
       }
   
       void addAtIndex(int index, int val) {
           int count = 0;
           LinkedNode *cur = _dummyHead;
           //遍历，比对下标
           while(count != index && cur){
               cur = cur->next;
               count++;
           }
           if(count != index){
               //如果遍历完了，仍然没找到索引，说明超标了
               return;
           }else if(cur){
               LinkedNode *node = new LinkedNode(val);
               node->val = val;
               node->next = cur->next;
               cur->next = node;
           }
       }
   
       void deleteAtIndex(int index) {
           int count = 0;
           LinkedNode *cur = _dummyHead;
           while(cur){
               if(count == index){
                   break;
               }else{
                   cur = cur->next;
                   count++;
               }
           }
           if(count == index && cur && cur->next){
               LinkedNode *delNode = cur->next;
               cur->next = delNode->next;
               delete delNode;
           }
           return;
       }
   private:
       LinkedNode* _dummyHead;
   };
   
   /**
    * Your MyLinkedList object will be instantiated and called as such:
    * MyLinkedList* obj = new MyLinkedList();
    * int param_1 = obj->get(index);
    * obj->addAtHead(val);
    * obj->addAtTail(val);
    * obj->addAtIndex(index,val);
    * obj->deleteAtIndex(index);
    */
   

6. 学习收获

   • 关于指针：使用前必须判断该指针是否为空

3.LeetCode 206.反转链表

1. 题目

   • https://leetcode.cn/problems/reverse-linked-list/

2. 思路

   • 建一个revHead没
   • 按head链表顺序读，读一个移动一个节点到revHead后，头插法
   • head后移一个

3. 刷随想录后想法

   • 双指针
   • 递归法

4. 实现困难

   • 递归

5. 实现代码

   /**
    * 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* reverseList(ListNode* head) {
           ListNode* revHead = new ListNode(0);
           revHead->next = nullptr;
           ListNode* cur;
           while(head){
               cur = head;
               head = head->next;
               cur->next = revHead->next;
               revHead->next = cur;
           }
           return revHead->next;
           }
   };
   

6. 学习收获