1、单链表逆序
思路图
代码实现
//著: 链表结构里记得加 friend void ReverseLink(Clink& link);
void ReverseLink(Clink& link)
{
Node* p = link.head_->next_;
while( p == nullptr)
{
return;
}
Node* q = p->next_;
link.head_->next_ = nullptr;
while(p != nullptr)
{
Node* q = p->next_;
//p指针指向的节点进行头插
p->next_ = link.head_->next_;
link.head_->next_ = p;
p = q;
}
/*
Node* p = link.head_->next_;
if( p == nullptr)
{
return;
}
Node* q = p->next_;
link.head_->next_ = nullptr;
while( q != nullptr)
{
link.head_->data_ = p->data_;
p->next_ = link.head_;
link.head_ = p;
p = q;
q = p->next_;
}
link.head_->data_ = p->data_;
p->next_ = link.head_;
link.head_ = p;
p = nullptr;
*/
}
测试
int main()
{
Clink link;
Clink link2;
srand(time(0));
for(int i = 0; i<10; i++)
{
int val = rand() % 100;
link.InsertTail(val);
}
link.show();
cout << endl;
ReverseLink(link);
ReverseLink(link2);
link.show();
cout << endl;
link2.show();
cout << endl;
}
运行结果
2、单链表倒数第k个节点
思路图
双指针同步位移，两个指针相聚k个节点
代码实现
//求倒数第k个节点的值
bool MyGetLastKNode(Clink& link,int k,int& val)
{
if(k<1)
{
return 0;
}
Node* p = link.head_;
if( p == nullptr)
{
return false;
}
Node* pre = link.head_;
for(int i = 0; i < k ; i++)
{
pre = pre->next_;
if( pre == nullptr )
{
return false;
}
}
//p在头节点，pre在正数第k个节点
while( pre != nullptr )
{
p = p->next_;
pre = pre->next_;
}
val = p->data_;
return true;
}
代码测试
int main()
{
Clink link;
Clink link2;
srand(time(0));
for(int i = 0; i<10; i++)
{
int val = rand() % 100;
link.InsertTail(val);
}
link.show();
cout << endl;
int val=0;
if(MyGetLastKNode(link,3,val))
{
cout<< "k == 3 ";
cout<< "find " << val<< endl;
}
else
{
cout<< "k == 3 ";
cout<< "false" << endl;
}
if(MyGetLastKNode(link,0,val))
{
cout<< "k == 0 ";
cout<< "find " << val<< endl;
}
else
{
cout<< "k == 0 ";
cout<< "false" << endl;
}
if(MyGetLastKNode(link,12,val))
{
cout<< "k == 12 ";
cout<< "find " << val<< endl;
}
else
{
cout<< "k == 12 ";
cout<< "false" << endl;
}
if(MyGetLastKNode(link2,3,val))
{
cout<< "k2 == 3 ";
cout<< "find " << val<< endl;
}
else
{
cout<< "k2 == 3 ";
cout<< "false" << endl;
}
}
运行结果
3、并两个有序单链表
思路图
代码实现
//合并两个有序单链表
bool MergeLink(Clink& link1,Clink& link2)
{
Node* p = link1.head_->next_;
Node* q = link2.head_->next_;
Node* last = link1.head_;
link2.head_->next_ = nullptr;
while(p != nullptr && q != nullptr)
{
if(p->data_ < q->data_)
{
last->next_ = p;
p = p->next_;
last = last->next_;
}
else
{
last->next_ = q;
q = q->next_;
last = last->next_;
}
}
if(p != nullptr)
{
last->next_ = p;
}
else
{
last->next_ = q;
}
/*
Node* pre = link1.head_;
Node* p = link1.head_->next_;
Node* q = link2.head_->next_;
while(q != nullptr)
{
if(p == nullptr && q != nullptr)
{
pre->next_ = q;
link2.head_->next_ = nullptr;
return true;
}
else if(p->data_ <= q->data_)//这里假设从小到大
{
p = p->next_;
pre = pre->next_;
}
else
{
link2.head_->next_ = q->next_;
pre->next_=q;
q->next_ = p;
pre=q;
q = link2.head_->next_;
}
}
*/
return true;
}
运行结果
4、判断单链表是否存在环以及入口节点
思路图
代码实现
//判断单链表是否存在环以及入口节点
//这里参数为Node,方便测试
//记得 friend
bool IsLinkHasCirle(Node* head,int& val)
{
Node* fast = head;
Node* slow = head;
while(fast != nullptr && fast->next_ != nullptr)
{
slow = slow->next_;
fast = fast->next_->next_;
if(slow == fast)
{
//快慢指针再次相遇，链表存在环
fast = head;
while(fast != slow)
{
slow = slow->next_;
fast = fast->next_;
}
val = slow->data_;
return true;
}
}
return false;
}
测试
int main()
{
Node head;
Node n1(25),n2(61),n3(312),n4(118);
head.next_ = &n1;
n1.next_ = &n2;
n2.next_ = &n3;
n3.next_ = &n4;
n4.next_ = &n2;
int val;
if(IsLinkHasCirle(&head,val))
{
cout<< "链表存在环，环的入口节点是： "<< val << endl;
}
}
运行结果
5、判断两个链表是否相交
思路图
代码实现
//判断两个链表是否相交,如果相交，返回相交节点的值
bool IsLinkHasMerge(Node* head1,Node* head2,int &val)
{
int cnt1 = 0,cnt2 = 0;
Node* p = head1->next_;
Node* q = head2->next_;
//计算两个链表的长度
while(p != nullptr)
{
p = p->next_;
cnt1++;
}
while(q != nullptr)
{
q = q->next_;
cnt2++;
}
p = head1->next_;
q = head2->next_;
if(cnt1 > cnt2)
{
//第一个链表长
cnt1 = cnt1-cnt2;
while(cnt1 >0)
{
p = p->next_;
cnt1--;
}
}
else
{
//第二个链表长
cnt2 = cnt2 -cnt1;
while(cnt2 > 0)
{
q = q->next_;
cnt2--;
}
}
while(p != nullptr && q != nullptr)
{
if(p == q)
{
val = p->data_;
return true;
}
p = p->next_;
q = q->next_;
}
return false;
}
测试代码
int main()
{
Node head1;
Node head2;
Node n1(25),n2(61),n3(312),n4(118),n5(18);
Node nn1(1),nn2(2);
head1.next_ = &n1;
n1.next_ = &n2;
n2.next_ = &n3;
n3.next_ = &n4;
n4.next_ = &n5;
head2.next_ = &nn1;
nn1.next_ = &nn2;
nn2.next_ = &n3;
int val;
if(IsLinkHasMerge(&head1,&head2,val))
{
cout<< "两个链表相交，相交的节点是： "<< val << endl;
}
}
运行结果
6、删除链表倒数第N个节点
思维图
代码实现
//这里使用利扣刷题
/**
* 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* removeNthFromEnd(ListNode* head, int n) {
//在函数内部给链表增加一个头节点，以解决不带头节点的单链表
//head_->next head
ListNode* head_ = new ListNode(0,head);
ListNode* first = head;
ListNode* second = head_;
for(int i = 0; i < n; i++)
{
first = first->next;
}
while(first)
{
first = first->next;
second = second->next;
}
second->next = second->next->next;
ListNode* ans = head_->next;
delete head_;
return ans;;
}
};
7、旋转链表
思路图
代码实现
/**
* 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* rotateRight(ListNode* head, int k) {
ListNode*p = head;
ListNode*q = head;
if(head == nullptr || k == 0)
{
return head;
}
int number = 0;//判断链表的长度
for(ListNode *k = head; k != nullptr; k = k->next)
{
number++;
}
k = k%number;
for(int i = 0; i < k; i++)
{
p = p->next;
}
while(p->next != nullptr)
{
q = q->next;
p = p->next;
}
p->next = head;
head = q->next;
q->next = nullptr;
return head;
}
};
