数据结构——单链表2

单链表2
单链表逆序问题
求单链表倒数第K个节点
合并两个有序单链表
判断单链表是否存在环以及入口节点

#include <iostream>
#include<stdlib.h>
#include<time.h>
using namespace std;

//节点类型
struct Node {
	Node(int data = 0) :data_(data), next_(nullptr) {}
	//节点类型的构造函数
	int data_;
	Node* next_;
};

//单链表代码实现
//只有在找尾节点指针才指向下一个
class Clink {
public:
	Clink() {
		//给head_初始化指向指向头节点
		head_ = new Node();
	}
	~Clink() {
		//delete []head_;删除数组类指针[]指针
		//head_=nullptr;
		//链表遍历每一个节点
		Node* p = head_;
		while (p != nullptr) {
			head_ = head_->next_;
			delete p;
			p = head_;
		}
		head_ = nullptr;
	}
public:
	//单链表尾插法  O(n)
	//若head_:头节点  tail_:尾节点  O(1)
	void insertTail(int val) {
		//先找的当前链表的末尾节点
		Node* p = head_;
		while (p->next_ != nullptr) {
			p = p->next_;
		}
		//生成新的节点
		Node* node = new Node(val);
		//把生成的新节点挂在尾节点后面
		p->next_ = node;
	}
	//单链表头插法
	void insertHead(int val) {//O(1)
		Node* node = new Node(val);
		node->next_ = head_->next_;
		head_->next_ = node;
	}
	
	//单链表删除多个节点
	void RemoveAll(int val) {
		Node* q = head_;
		Node* p = head_->next_;
		while (p != nullptr) {
			if (p->data_ == val) {
				q->next_ = p->next_;
				delete p;
				//对指针p进行重置
				p = q->next_;
			}
			else {
				q = p;
				p = p->next_;
			}
		}
	}

	//链表的打印
	void Show() {//从第一个节点打印到最后一个节点
		Node* p = head_->next_;
		while (p != nullptr) {
			cout << p->data_ << " ";
			p = p->next_;
		}
		cout << endl;
	}
private:
	Node* head_;//指向链表的头节点

	friend void ReverseLink(Clink& link);
	friend bool GetLastkKNode(Clink& link, int k, int& val);
	friend void MergeLink(Clink& link1, Clink& link2);
};

//c接口全局接口
//单链表逆序
void ReverseLink(Clink &link) {
	Node* p = link.head_->next_;
	if (p == nullptr) {
		return;
	}
	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;
	}
}

//求倒数第k个节点的值
bool GetLastKNode(Clink& link, int k, int& val) {
	Node* head = link.head_;
	Node* pre = head;
	Node* p = head;

	if (k < 1) {
		return false;
	}
	for (int i = 0; i < k; i++) {
		p = p->next_;
		if (p == nullptr) {
			return  false;
		}
	}
	//pre在头节点，p在正数第k个节点
	while (p != nullptr) {
		pre = pre->next_;
		p = p->next_;
	}
	val = pre->data_;
	return true;
}

//合并两个有序的单链接
void 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;
	}

}
//是否存在环
bool IsLinkHasCircle(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;
}

int main() {
    Clink link;
	srand(time(0));
	for (int i = 0; i < 10; i++) {
		int val = rand() % 100;
		link.insertTail(val);
	}
	link.Show();

	ReverseLink(link);
	link.Show();

	int kval;
	int k = 3;
	if (GetLaskKNode(link, 3, kval)) {
		cout << "倒数第" << k << "个节点的值：" << kval << endl;
	}


	int arr[] = { 25,37,52,78,88,92,98,108 };
	int brr[] = { 13,23,40,56,62,77,109 };

	Clink link1;
	Clink link2;

	for (int v : arr) {
		link1.insertTail(v);
	}
	for (int v : brr) {
		link2.insertTail(v);
	}
	link1.Show();
	link2.Show();

	MergeLink(link1, link2);
	link1.Show();

Node head;
	Node n1(25), n2(67), n3(32), n4(18);
	head.next_ = &n1;
	n1.next_ = &n2;
	n2.next_ = &n3;
	n3.next_ = &n4;
	n4.next_ = &n2;

	int val;
	if (IsLinkHasCircle(&head,val)) {
		cout << "链表存在环，环的入口节点是：" << val << endl;
	}



}