1. 删除链表中等于给定值 val 的所有结点。
203. 移除链表元素 - 力扣（LeetCode）
方法一：
struct ListNode* removeElements(struct ListNode* head, int val)
{
struct ListNode* prev=NULL;
struct ListNode* cur=head;
while(cur)
{
if(cur->val==val)
{
struct ListNode* next = cur->next;
free(cur);
if(prev)
prev->next=next;
else
head = next;
cur=next;
}
else
{
prev=cur;
cur=cur->next;
}
}
return head;
}
方法二：
哨兵位的头节点不存储有效数据
struct ListNode* removeElements(struct ListNode* head, int val)
{
struct ListNode* newhead = NULL,*tail=NULL;
struct ListNode* cur=head;
//哨兵位
newhead =tail=(struct ListNode*)malloc(sizeof(struct ListNode));
while(cur)
{
//不是val的节点拿下来尾插
if(cur->val !=val)
{
tail->next=cur;
tail=tail->next;
cur=cur->next;
}
else
{
struct LIstNode* tmp= cur;
cur=cur->next;
free(tmp);
}
}
tail->next =NULL;
struct ListNode* tmp =newhead;
newhead=newhead->next;
free(tmp);
return newhead;
}
2.反转一个单链表。
206. 反转链表 - 力扣（LeetCode）
struct ListNode* reverseList(struct ListNode* head)
{
if(head==NULL)
retuen NULL;
struct ListNode* n1 ,*n2.n3;
n1 = NULL;
n2 = head;
n3 = head->next;
while(n2)
{
n2->next=n1;
n1=n2;
n2=n3
if(n3)
n3=n3->next;
}
return n1;
}
3. 给定一个带有头结点 head 的非空单链表，返回链表的中间结点。如果有两个中间结点，则 返回第二个中间结点。
876. 链表的中间结点 - 力扣（LeetCode）
struct ListNode* middleNode(struct ListNode* head)
{
struct ListNode*prev=head,*tail=head;
while(tail && tail->next)
{
prev=prev->next;
tail=tail->next->next;
}
return prev;
}
4. 输入一个链表，输出该链表中倒数第k个结点。
面试题 02.02. 返回倒数第 k 个节点 - 力扣（LeetCode）
int kthToLast(struct ListNode* head, int k)
{
struct ListNode*cur=head;
while(cur)
{
if(cur->val==k)
{
return cur->val;
}
cur=cur->next;
}
return NULL;
}
5. 将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有 结点组成的。
21. 合并两个有序链表 - 力扣（LeetCode）
typedef struct ListNode ListNode;
struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2)
{
if(list1==NULL)
return list2;
if(list2==NULL)
return list1;
ListNode* phead1 ,*phead2 ;
ListNode* phead3 ,*cur;
phead3 = cur = (ListNode*)malloc(sizeof(ListNode));
phead1 = list1;
phead2 = list2;
while(phead1 && phead2)
{
if(phead1->val <= phead2->val)
{
cur->next = phead1;
phead1 = phead1->next;
}
else
{
cur->next = phead2;
phead2 = phead2->next;
}
cur = cur->next;
}
if(phead1)
{
cur->next = phead1;
}
if(phead2)
{
cur->next = phead2;
}
phead1 = phead3->next;
free(phead3);
return phead1;
}
6. 编写代码，以给定值x为基准将链表分割成两部分，所有小于x的结点排在大于或等于x的结点之前 。
链表分割_牛客题霸_牛客网 (nowcoder.com)
class Partition {
public:
ListNode* partition(ListNode* pHead, int x)
{
// write code here
struct ListNode *head1,*tail1,*head2,*tail2;
head1 = tail1 = (struct ListNode*) malloc(sizeof(struct ListNode));
head2 = tail2 = (struct ListNode*) malloc(sizeof(struct ListNode));
struct ListNode *cur = pHead;
while(cur)
{
if(cur->val <x)
{
tail1->next = cur;
tail1 = tail1->next;
}
else
{
tail2->next = cur;
tail2 = tail2->next;
}
cur=cur->next;
}
tail1->next = head2->next;
tail2->next = NULL;
pHead = head1->next;
free(head1);
free(head2);
return pHead;
}
};
7. 链表的回文结构。
链表的回文结构_牛客题霸_牛客网 (nowcoder.com)
class PalindromeList {
public:
struct ListNode* reverseList(struct ListNode* head)
{
if(head == NULL)
return NULL;
struct ListNode* n1 ,*n2,*n3;
n1 = NULL;
n2 = head;
n3 = head->next;
while(n2)
{
n2->next=n1;
n1=n2;
n2=n3;
if(n3)
n3=n3->next;
}
return n1;
}
struct ListNode* middleNode(struct ListNode* head)
{
struct ListNode*prev=head,*tail=head;
while(tail && tail->next)
{
prev=prev->next;
tail=tail->next->next;
}
return prev;
}
bool chkPalindrome(ListNode* head)
{
// write code here
struct ListNode* mid = middleNode(head);
struct ListNode* rhead = reverseList(head);
while(head && rhead)
{
if(head->val != rhead->val)
{
return false;
}
head = head->next;
rhead = rhead ->next;
}
return true;
}
};
思路：先找中间，再逆置，然后比较
8. 环形链表I
141. 环形链表 - 力扣（LeetCode）
bool hasCycle(struct ListNode *head)
{
struct ListNode * slow=head,*fast=head;
while(fast && fast->next)
{
fast=fast->next->next;
slow=slow->next;
if(fast== slow)
return true;
}
return false;
}
9.环形链表II
142. 环形链表 II - 力扣（LeetCode）
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode *detectCycle(struct ListNode *head)
{
struct ListNode * slow=head,*fast=head;
while(fast && fast->next)
{
slow=slow->next;
fast=fast->next->next;
if(fast==slow)
{
struct ListNode * meet = slow;
while(head!=meet)
{
head=head->next;
meet=meet->next;
}
return meet;
}
}
return NULL;
}
10. 链表复制
138. 随机链表的复制 - 力扣（LeetCode）
struct Node* copyRandomList(struct Node* head)
{
struct Node* cur=head;
while(cur)
{
struct Node* copy=(struct Node*)malloc(sizeof(struct Node));
copy->val = cur->val;
copy->next = cur->next;
cur->next = copy;
cur = copy->next;
}
cur = head;
while(cur)
{
struct Node* copy = cur->next;
if(cur->random == NULL)
{
copy->random = NULL;
}
else
{
copy->random = cur->random->next;
}
cur = copy->next;
}
struct Node*newhead=NULL,*tail = NULL;
cur = head;
while(cur)
{
struct Node* copy = cur->next;
struct Node* next = copy->next;
if(tail == NULL)
{
newhead = tail =copy;
}
else
{
tail->next = copy;
tail = tail->next;
}
cur->next = next;
cur=next;
}
return newhead;
}
11. 输入两个链表，找出它们的第一个公共结点。
160. 相交链表 - 力扣（LeetCode）
struct ListNode *getIntersectionNode(struct ListNode* headA, struct ListNode* headB)
{
struct ListNode *curA=headA,*curB=headB;
int lenA=1,lenB=1;
while(curA)
{
curA=curA->next;
lenA++;
}
while(curB)
{
curB=curB->next;
lenB++;
}
if(curA!=curB)
return NULL;
int n=abs(lenA-lenB);
struct ListNode* longlist=headA,*shortlist = headB;
if(lenA<lenB)
{
longlist = headB;
shortlist = headA;
}
while(n--)
{
longlist=longlist->next;
}
while(longlist!=shortlist)
{
longlist=longlist->next;
shortlist=shortlist->next;
}
return shortlist;
}
