• 203.移除链表元素
• 206.反转链表
• 160.相交链表
• 707.设计链表
• 24.两两交换链表中的节点
• 19.删除链表的倒数第 N 个结点
• 142.环形链表 II

• ListNode结构体要会背

struct ListNode {int val;ListNode* next;ListNode(int _val = 0, ListNode* _next = nullptr): val(_val), next(_next){}
};
// 结尾记得加分号

• 先判断head节点是否为nullptr
• 先判断有效性，再取next
• 善用dummy哨兵节点和pre
• 使用pre，cur，next，next2指代已有节点，使用temp指代新建节点
• 前向迭代模板

ListNode dummy = ListNode(-1, head);
ListNode *cur = dummy, *next;
for (next = cur->next; cur && next; next = cur->next) {...cur = cur->next;
}

• 快慢指针，记得先判断快指针有效性，再取next
• 相向指针
• 相交链表，尾部对齐，再逐个检查尾部指针，相同处为相交处
• 设计链表时，不要单独实现头插和尾插，统一用指定下标插入实现
• 循环链表，两步指针与单步指针重合处，必定为head
• 有环链表，在快慢指针重合处、head，同时出发，重合点为环的起点

• 707.设计链表

class MyLinkedList {struct ListNode {int val;ListNode* next;ListNode(int _val = 0, ListNode* _next = nullptr): val(_val), next(_next){}};private:ListNode* dummy;int size;public:MyLinkedList(): dummy(new ListNode()), size(0){}int get(int index){if (index >= size || index < 0)return -1;ListNode *cur = dummy, *next;int cur_index = 0;for (next = cur->next; cur && next; next = cur->next) {if (cur_index == index) {return next->val;}cur = cur->next;++cur_index;}return cur->val;}// 不要单独实现头插和尾插，统一用指定下标插入实现void addAtHead(int val) { addAtIndex(0, val); }void addAtTail(int val) { addAtIndex(size, val); }void addAtIndex(int index, int val){if (index > size)return;if (index < 0)index = 0;ListNode *cur = dummy, *next, *temp;int cur_index = 0;for (next = cur->next; cur; next = cur->next) {if (cur_index == index) {temp = new ListNode(val);temp->next = next;cur->next = temp;++size;break;}cur = cur->next;++cur_index;}}void deleteAtIndex(int index){if (index >= size)return;ListNode *cur = dummy, *next, *temp;int cur_index = 0;for (next = cur->next; cur && next; next = cur->next) {if (cur_index == index) {temp = next;cur->next = next->next;delete temp;--size;break;}cur = cur->next;++cur_index;}}
};

• 24.两两交换链表中的节点

class Solution {
public:ListNode* swapPairs(ListNode* head){if (!head || !(head->next))return head;ListNode dummy = ListNode(-1, head);ListNode *pre = &dummy, *cur = dummy.next, *next, *next2;for (next = cur->next; cur && next; next = cur->next) {next2 = next->next;pre->next = next;cur->next = next2;next->next = cur;// 易错点：pre要先更新，并且pre=curpre = cur;cur = next2;// 因为cur=next2，跳过了next// 所以更新后的cur无效，则结束，否则next的有效性无法确保if (!cur)break;}return dummy.next;}
};

• 142.环形链表 II

class Solution {
public:ListNode* detectCycle(ListNode* head){if (!head)return head;if (!(head->next))return head->next;ListNode *slow, *fast;// 易错点：快慢指针有效性判断for (slow = fast = head;slow->next && fast->next && fast->next->next;) {slow = slow->next;fast = fast->next->next;if (slow == fast) {for (; slow != head;) {slow = slow->next;head = head->next;}return head;}}return nullptr;}
};