day3 补

Leetcode 203 删除链表节点

感觉使用虚拟头节点方便理解，如果不使用虚拟头节点需要单独写一段处理头节点的代码

class Solution {
public:ListNode* removeElements(ListNode* head, int val) {ListNode* dummyHead = new ListNode(0); // 设置一个虚拟头结点dummyHead->next = head; // 将虚拟头结点指向head，这样方便后面做删除操作ListNode* cur = dummyHead;while (cur->next != NULL) {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;}
};

Leetcode 707设计链表

class MyLinkedList {
public:struct LinkNode{LinkNode*prev;LinkNode*next;int val_;LinkNode(int val):val_(val),prev(nullptr),next(nullptr){};};MyLinkedList() {dummyHead=new LinkNode(0);dummyTail=new LinkNode(0);size_=0;dummyHead->next=dummyTail;dummyTail->prev=dummyHead;}int get(int index) {if(index>=size_||index<0)return -1;int num;LinkNode*cur=dummyHead;for(int i=0;i<index;++i){cur=cur->next;}num=cur->next->val_;delete cur;return num;}void addAtHead(int val) {LinkNode*newNode=new LinkNode(val);LinkNode*temp=new LinkNode(0);temp->next=dummyHead->next;dummyHead->next=newNode;newNode->prev=dummyHead;newNode->next=temp->next;temp->next->prev=newNode;size_++;delete temp;}void addAtTail(int val) {LinkNode*newNode=new LinkNode(val);LinkNode*temp=new LinkNode(0);temp->prev=dummyTail->prev;dummyTail->prev=newNode;newNode->next=dummyTail;newNode->prev=temp->prev;temp->prev->next=newNode;size_++;delete temp;}void addAtIndex(int index, int val) {if(index==0){addAtHead(val);return ;}if(index==size_){addAtTail(val);return;}if(index>size_){return;}LinkNode*cur=dummyHead;LinkNode*newNode=new LinkNode(val);LinkNode*temp=new LinkNode(0);for(int i=0;i<index;i++){cur=cur->next;}temp->next=cur->next;cur->next=newNode;newNode->prev=cur;newNode->next=temp->next;temp->next->prev=newNode;size_++;delete temp;}void deleteAtIndex(int index) {if(index>=size_||index<0)return;LinkNode*cur=dummyHead;LinkNode*temp=new LinkNode(0);for(int i=0;i<index;i++){cur=cur->next;}temp->next=cur->next->next;cur->next=cur->next->next;temp->next->prev=cur;size_-=1;delete temp->next;delete temp;}private:LinkNode*dummyHead;LinkNode*dummyTail;int size_;
};/*** 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);*/

如上面所示代码，有问题，但是看不出来哪有问题。随后按照代码随想录的思路进行编写，其中最需要注意的位置是get、addAtIndex、deleteAtIndex中的循环条件，其中get需要cur正好位于index，而add、delete需要cur位于index位置的前一个或者后一个，这道题需要多练几遍。

class MyLinkedList {
public:struct LinkNode{LinkNode* prev;LinkNode* next;int val_;LinkNode(int val) :val_(val), prev(nullptr), next(nullptr) {};};MyLinkedList() {sentinelNode = new LinkNode(0);size_ = 0;sentinelNode->next = sentinelNode;sentinelNode->prev = sentinelNode;}int get(int index) {if (index >= size_ || index < 0)return -1;int num;LinkNode* cur = sentinelNode;int mid = size_ >> 1;if (index < mid){for (int i = 0; i < index -1; i++){cur = cur->next;}}else {for (int i = 0; i < size_ - index; i++){cur = cur->prev;}}num = cur->val_;return num;}void addAtHead(int val) {LinkNode* newNode = new LinkNode(val);LinkNode* temp = sentinelNode->next;newNode->prev = sentinelNode;newNode->next = sentinelNode->next;sentinelNode->next = newNode;temp->prev = newNode;size_++;}void addAtTail(int val) {LinkNode* newNode = new LinkNode(val);LinkNode* temp;temp = sentinelNode->prev;newNode->next = sentinelNode;newNode->prev = temp;sentinelNode->prev = newNode;temp->next = newNode;size_++;}void addAtIndex(int index, int val) {if (index == 0){addAtHead(val);return;}if (index > size_){return;}int mid = size_ >> 1;LinkNode* cur = sentinelNode;LinkNode* newNode = new LinkNode(val);if (index < mid){for (int i = 0; i < index; i++){cur = cur->next;}LinkNode* temp = cur->next;cur->next = newNode;temp->prev = newNode;newNode->next = temp;newNode->prev = cur;size_++;}else{for (int i = 0; i < size_ - index ; ++i){cur = cur->prev;}LinkNode* temp = cur->prev;cur->prev = newNode;temp->next = newNode;newNode->prev = temp;newNode->next = cur;size_++;}}void deleteAtIndex(int index) {if (index >= size_ || index < 0)return;int mid = size_ >> 1;LinkNode* cur = sentinelNode;if (index < mid){for (int i = 0; i < index; ++i){cur = cur->next;}LinkNode* temp = cur->next->next;cur->next = temp;temp->prev = cur;size_--;}else {for (int i = 0; i < size_ - index - 1; ++i){cur = cur->prev;}LinkNode* temp = cur->prev->prev;cur->prev = temp;temp->next = cur;size_--;}}private:LinkNode* sentinelNode;int size_;
};

Leetcode 206

单向链表反转，之前没接触过，应该说单向链表都没怎么接触过，对于我眼前一亮的地方在于

ListNode*pre=nullptr;

这样cur->next直接=pre即可把尾节点next赋值为nullptr

判断条件为cur，当cur为nullptr时pre正好成为新的head

通过画图可以很好理解

class Solution {
public:ListNode* reverseList(ListNode* head) {ListNode*cur=head;ListNode*prev=nullptr;while(cur){ListNode*temp=cur->next;cur->next=prev;prev=cur;cur=temp;}return prev;}
};

很少写链表，更别说设计链表了，画图画了半天有点懵，中间边界条件判断很难看出来错误，还是需要多写，下周需要复习这一部分内容，通过画图能够更加直观了解问题。