代码随想录算法训练营第三天| 203.移除链表元素 、 707.设计链表 、206.反转链表
链表的构造:
link.h:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #ifndef LINK_H 2 #define LINK_H 3 #include<vector> 4 5 struct ListNode { 6 int val; 7 ListNode* next; 8 ListNode() :val(0), next(nullptr) {} 9 ListNode(int x) : val(x), next(nullptr) {} 10 ListNode(int x, ListNode* next) : val(x), next(next) {} 11 }; 12 13 class Solution { 14 public: 15 ListNode* removeElements(ListNode* head, int val); 16 17 }; 18 #endif
实现链表:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include"Link.h" 3 using namespace std; 4 5 ListNode* Solution::removeElements(ListNode* head, int val) 6 {}
实现移除元素:
难点:
1,因为给的head是不是空节点的head,一定要注意这一点
2,需要自己定义一个空的头节点
3,需要设置一个游标
4,返回的时候,返回空节点的时候->用num
1 ListNode* Solution::removeElements(ListNode* head, int val) 2 { 3 ListNode* DyHead = new ListNode(); 4 //注意 cur_ == head 5 DyHead->next = head; 6 auto cur_ = DyHead; 7 8 while (cur_->next != nullptr) 9 { 10 if (cur_->next->val == val) 11 { 12 ListNode* mid = cur_->next; 13 cur_->next = mid->next; 14 } 15 // 不是的话,就下一个 16 else 17 { 18 cur_ = cur_->next; 19 } 20 21 } 22 23 //不是原数组的head 而是 dyhead->next 24 return DyHead->next; 25 }
设计列表 ,仍然出现访问空指针的问题,需要查看
class MyLinkedList { public: struct LinkNode { int val; LinkNode* next; LinkNode(int v) : val(v), next(nullptr) {}; }; MyLinkedList() { head = new LinkNode(0); head->next = nullptr; } int get(int index) { auto cur_ = head; int idx = 0; while (cur_->next != nullptr) { if (idx == index) { return cur_->next->val; } else { cur_ = cur_->next; } idx++; } return -1; } void addAtHead(int val) { auto* cur = head->next; LinkNode* mid = new LinkNode(val); head->next = mid; mid->next = cur; } void addAtTail(int val) { LinkNode* mid = new LinkNode(val); auto* cur = head; while (cur->next != nullptr) { cur = cur->next; } cur->next = mid; } void addAtIndex(int index, int val) { int length = 0, idx = 0; LinkNode* cur = head; LinkNode* newNode = new LinkNode(val); // 一定要把循环写进去,因为while(index--),最后一层,会把index -- 多一步 while (index > 0) { if (cur->next != nullptr) { cur = cur->next; } else { break; } index = index - 1; } if (index == 0 && cur->next != nullptr) { LinkNode* mid = cur->next; cur->next = newNode; newNode->next = mid; } else if (index == 0 && cur->next == nullptr) { cur->next = newNode; } } void deleteAtIndex(int index) { LinkNode* cur = head; while (index >0) { if (cur->next != nullptr) { cur = cur->next; } else { break; } index = index - 1; } if (index == 0) { cur->next = cur->next->next; } } private: LinkNode* head; }; /** * 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); */
反转列表:
1 ListNode* MyLinkedList::reverseList(ListNode* head) { 2 ListNode* dyHead = new ListNode(); 3 dyHead->next = head; 4 auto* cur = dyHead; 5 6 ListNode* result = new ListNode(); 7 while (cur->next !=nullptr) 8 { 9 ListNode* currentNode = new ListNode(cur->next->val); 10 auto mid = result->next; 11 result->next = currentNode; 12 currentNode->next = mid; 13 14 cur = cur->next; 15 } 16 17 return result->next; 18 }