您现在的位置是:首页 >技术杂谈 >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。网站首页技术杂谈
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。
简介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。
LeetCode 203 移除链表元素
题目链接 203.移除链表元素
/**
* 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);//创建虚拟头节点,使对所有节点的删除一致
dummyhead->next = head;
ListNode* cur = dummyhead;
while(cur->next != NULL){
if(cur->next->val == val){
ListNode* tmp = cur->next;
cur->next = tmp->next;
delete tmp;
}else
cur = cur->next;
}
head = dummyhead->next;
delete dummyhead;
return head;
}
};
看到题的第一想法
建立虚拟头结点,那么所有结点的删除操作一致
LeetCode 707 设计链表
题目链接 707.设计链表
class MyLinkedList {
public:
struct LinkNode{
int val;
LinkNode *next;
LinkNode():val(0),next(nullptr){}
LinkNode(int x):val(x),next(nullptr){}
LinkNode(int x,LinkNode *next):val(x),next(next){}
};
MyLinkedList() {
_dummyhead = new LinkNode(0);
_size = 0;
}
int get(int index) {
if(index < 0 || index > _size - 1)
return -1;
LinkNode* cur = _dummyhead->next;
while(index--){
cur = cur->next;
}
return cur->val;
}
void addAtHead(int val) {
LinkNode* cur = new LinkNode(val);
cur->next = _dummyhead->next;
_dummyhead->next = cur;
++_size;
}
void addAtTail(int val) {
LinkNode* newnode = new LinkNode(val);
LinkNode* cur = _dummyhead;
while(cur->next!=nullptr)
cur = cur->next;
newnode->next = cur->next;
cur->next = newnode;
++_size;
}
void addAtIndex(int index, int val) {
if(index < 0 || index > _size)
return;
LinkNode* newnode = new LinkNode(val);
LinkNode* cur = _dummyhead;
while(index--)
cur = cur->next;
newnode->next = cur->next;
cur->next = newnode;
++_size;
}
void deleteAtIndex(int index) {
if(index < 0 || index > _size - 1)
return;
LinkNode* cur = _dummyhead;
while(index--)
cur = cur->next;
LinkNode* delnode = cur->next;
cur->next = delnode->next;
delete delnode;
--_size;
}
private:
int _size;
LinkNode* _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);
*/
看到题的第一想法
直接码
遇到的困难
- 对于private和public的理解不到位
看完代码随想录后的想法
- 在private里定义_size和_dummynode
LeetCode 206 反转链表
题目链接 206.反转链表
/**
* 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* dummynode = new ListNode(0);
dummynode->next = head;
ListNode* cur = dummynode->next;
ListNode* latter;
while(cur->next != nullptr){
latter = cur->next;
cur->next = latter->next;//尾插法
latter->next = dummynode->next;
dummynode->next = latter;
}
head = dummynode->next;
delete dummynode;
return head;
}
};
看到题的第一想法
头插法
遇到的困难
- Line 18: Char 20: runtime error: member access within null pointer of type ‘ListNode’ (solution.cpp)
SUMMARY: UndefinedBehaviorSanitizer: undefined-behavior prog_joined.cpp:27:20
//头插法
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if(head==nullptr)
return nullptr;
//虚拟头结点
ListNode* dummynode = new ListNode(0);
dummynode->next = head;
ListNode* cur = dummynode->next;
ListNode* latter;
while(cur->next){
latter = cur->next;
cur->next = latter->next;//尾插法
latter->next = dummynode->next;
dummynode->next = latter;
}
head = dummynode->next;
delete dummynode;
return head;
}
};
//双指针
class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode *temp;
ListNode *cur = head;
ListNode *pre = nullptr;
while(cur){
temp = cur->next;
cur->next = pre;
pre = cur;
cur = temp;
}
return pre;
}
};
//递归
class Solution {
public:
ListNode* reverseList(ListNode* head) {
return reverse(nullptr,head);
}
ListNode* reverse(ListNode* pre,ListNode* cur){
if(cur == nullptr)
return pre;
ListNode* temp = cur->next;
cur->next = pre;
return reverse(cur,temp);
}
};
与第一次代码多了一次head==nullptr的判断,AC了。
看完代码随想录后的想法
- 还可以用双指针,递归的方法完成
风语者!平时喜欢研究各种技术,目前在从事后端开发工作,热爱生活、热爱工作。