左神基础班- 可能有环可能无环的链表 -判断是否相交

最新推荐文章于 2024-05-14 12:10:39 发布

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博客给出有环与无环链表不相交的代码组合，使用 return nullptr；表示该情况，还提及有完整带测试的代码。

//判断是否有环
LinkNode* loop(LinkNode* head){
	LinkNode* f = head->next;
	LinkNode* s = head->next;
	//注意这边的 &&
	while(f->next != nullptr && f->next->next != nullptr){
		s = s->next;
		f = f->next->next;
		if(s == f){
			f = head->next;
			break;
		}
	}
	//代表无环
	if(f->next==nullptr || f->next->next == nullptr){
		return nullptr;
	}
	//若有环
	while(s != f){
		s = s->next;
		f = f->next;
	}
	return s;
}

//2无环单链表是否相交
LinkNode* noloop(LinkNode* head1, LinkNode* head2){
	int num1 = 0;
	int num2 = 0;
	LinkNode* go1 = head1;
	LinkNode* go2 = head2;
	while(go1->next != nullptr){
		go1 = go1->next;
		num1++;
	}
	while(go2->next != nullptr){
		go2 = go2->next;
		num2++;
	}
	if(go1 != go2){
		return nullptr;
	}
	num1 = num1-num2;
	go1 = num1 > 0? head1:head2;
	go2 = go1 == head1? head2 : head1;
	if(num1<0){
		num1 = -num1;
	}
	while(num1 > 0){
		go1 = go1->next;
		num1--;	
	}
	while(go1 != go2){
		go1 = go1->next;
		go2 = go2->next;
	}
	return go1;
}

//两个有环链表 是否相交
LinkNode* hasloop(LinkNode* head1, LinkNode* head2){
	//在环外的链上相交,则其入环节点一定相同
	if(loop(head1) == loop(head2)){
		LinkNode* enloop = loop(head1);
		LinkNode* go1 = head1;
		LinkNode* go2 = head2;
		int num1 = 0;
		int num2 = 0;
		while(go1 ->next != enloop){
			num1++;
			go1 = go1->next;
		}
		while(go2->next != enloop){
			num2++;
			go2 = go2->next;
		}
		num1 = num1 - num2;
		go1 = num1 > 0? head1 : head2;
		go2 = go1 == head1 ? head2 : head1;
		if(num1<0)
			num1 = -num1;
		while(num1 > 0){
			go1 = go1->next;
			num1--;
		}
		while(go1 != go2){
			go1 = go1->next;
			go2 = go2->next;
		}
		return go1;
	}else{
		//另外两种结构： 66 或 三毛头没有中间一根毛
		//从一个入环节点开始沿着环遍历，如果重新回到原始位置没碰到另外一个入环节点，则为66结构
		LinkNode* enloop1 = loop(head1);
		LinkNode* enloop1_go = enloop1;
		LinkNode* enloop2 = loop(head2);
		while(enloop1_go->next != enloop1){
			enloop1_go = enloop1_go->next;
			if(enloop1_go == enloop2){
				return enloop1;
			}
		}
		//66结构
		return nullptr;
	}
}

最终组合起来使用： return nullptr；表示一个有环一个无环的链表比不相交

LinkNode* judge(LinkNode* head1, LinkNode* head2){
    if(head1 == nullptr || head2 == nullptr){
        return nullptr;
    }
	// 两个无环单链表是否相交
	if(loop(head1) == nullptr && loop(head2) == nullptr){
		return noloop(head1, head2);
	}else if(loop(head1) != nullptr && loop(head2)!= nullptr){
		return hasloop(head1, head2);
	}
	return nullptr;
}

完整的，带测试的代码：

//拷贝带有随机指针的单链表
#include<iostream>
#include<stack>
#include<unordered_map>
using namespace std;
struct LinkNode{
	int value;
	LinkNode* next;
	LinkNode(int a){
		value = a;
		next = nullptr;
	}
};
LinkNode* loop(LinkNode* head);
LinkNode* noloop(LinkNode* head1, LinkNode* head2);
LinkNode* hasloop(LinkNode* head1, LinkNode* head2);
LinkNode* judge(LinkNode* head1, LinkNode* head2);
int main(){
	LinkNode* head1 = new LinkNode(0);
	LinkNode* a = new LinkNode(1);
	LinkNode* b = new LinkNode(2);
	LinkNode* c = new LinkNode(3);
	LinkNode* d = new LinkNode(4);
	head1->next = a;
	a->next = b;
	b->next = c;
	c->next = d;
	//加环
	d->next = b;

	LinkNode* head2 = new LinkNode(0);
	LinkNode* a2 = new LinkNode(1);
	LinkNode* b2 = new LinkNode(2);
	LinkNode* c2 = new LinkNode(3);
	LinkNode* d2 = new LinkNode(4);
	head2->next = a2;
	a2->next = b2;
	b2->next = c2;
	c2->next = d2;
	d2->next = b2;

	// b2->next = c2;
	//判断一个单链表是否有环
	// if(loop(head2)){
	// 	cout << loop(head)->value <<endl;
	// }else{
	// 	cout << "无环" << endl;
	// }

	// 测试 66 结构
	if(judge(head1, head2) == nullptr){
		cout << "66"<< endl;
	}

	LinkNode* head3 = new LinkNode(0);
	LinkNode* a3 = new LinkNode(1);
	LinkNode* b3 = new LinkNode(2);
	LinkNode* c3 = new LinkNode(3);
	LinkNode* d3 = new LinkNode(4);
	LinkNode* e3 = new LinkNode(5);
	head3->next = a3;
	a3->next = b3;
	b3->next = c3;
	c3->next = d3;
	d3->next = e3;
	e3->next = c3;

	LinkNode* head4 = new LinkNode(0);
	LinkNode* a4 = new LinkNode(1);
	head4->next = a4;
	a4->next = b3;
	//测试环外相遇结构
	if(judge(head3,head4) != nullptr){
		cout << "seconde mode: the node is "<<judge(head3, head4)->value << endl;
	}
	
	LinkNode* head5 = new LinkNode(0);
	LinkNode* a5 = new LinkNode(1);
	head5->next = a5;
	a5->next = d3;
	//环内相交节点
	if(judge(head3,head5) != nullptr){
		cout <<"third mode: the node is"<< judge(head3, head5)->value << endl;	
	}
	

	LinkNode* head6 = new LinkNode(0);
	LinkNode* a6 = new LinkNode(1);
	LinkNode* b6 = new LinkNode(2);
	LinkNode* c6 = new LinkNode(3);
	LinkNode* d6 = new LinkNode(4);
	LinkNode* e6= new LinkNode(5);
	head6->next = a6;
	a6->next = b6;
	b6->next = c6;
	c6->next = d6;
	d6->next = e6;
	LinkNode* head7 = new LinkNode(0);
	LinkNode* a7 = new LinkNode(1);
	LinkNode* b7 = new LinkNode(2);
	head7->next = a7;
	a7->next = b7;
	b7->next = c6;
	//两单链表相交
	if(judge(head6,head7) != nullptr){
		cout <<"two Link have banana, the number : "<< judge(head6, head7)->value;
	}else{
		cout << "two Link no banana" <<endl;
	}
	return 0;
}

//判断是否有环
LinkNode* loop(LinkNode* head){
	LinkNode* f = head->next;
	LinkNode* s = head->next;
	//注意这边的 &&
	while(f->next != nullptr && f->next->next != nullptr){
		s = s->next;
		f = f->next->next;
		if(s == f){
			f = head->next;
			break;
		}
	}
	//代表无环
	if(f->next==nullptr || f->next->next == nullptr){
		return nullptr;
	}
	//若有环
	while(s != f){
		s = s->next;
		f = f->next;
	}
	return s;
}
//2无环单链表是否相交
LinkNode* noloop(LinkNode* head1, LinkNode* head2){
	int num1 = 0;
	int num2 = 0;
	LinkNode* go1 = head1;
	LinkNode* go2 = head2;
	while(go1->next != nullptr){
		go1 = go1->next;
		num1++;
	}
	while(go2->next != nullptr){
		go2 = go2->next;
		num2++;
	}
	if(go1 != go2){
		return nullptr;
	}
	num1 = num1-num2;
	go1 = num1 > 0? head1:head2;
	go2 = go1 == head1? head2 : head1;
	if(num1<0){
		num1 = -num1;
	}
	while(num1 > 0){
		go1 = go1->next;
		num1--;	
	}
	while(go1 != go2){
		go1 = go1->next;
		go2 = go2->next;
	}
	return go1;
}
//两个有环链表 是否相交
LinkNode* hasloop(LinkNode* head1, LinkNode* head2){
	//在环外的链上相交,则其入环节点一定相同
	if(loop(head1) == loop(head2)){
		LinkNode* enloop = loop(head1);
		LinkNode* go1 = head1;
		LinkNode* go2 = head2;
		int num1 = 0;
		int num2 = 0;
		while(go1 ->next != enloop){
			num1++;
			go1 = go1->next;
		}
		while(go2->next != enloop){
			num2++;
			go2 = go2->next;
		}
		num1 = num1 - num2;
		go1 = num1 > 0? head1 : head2;
		go2 = go1 == head1 ? head2 : head1;
		if(num1<0)
			num1 = -num1;
		while(num1 > 0){
			go1 = go1->next;
			num1--;
		}
		while(go1 != go2){
			go1 = go1->next;
			go2 = go2->next;
		}
		return go1;
	}else{
		//另外两种结构： 66 或 三毛头没有中间一根毛
		//从一个入环节点开始沿着环遍历，如果重新回到原始位置没碰到另外一个入环节点，则为66结构
		LinkNode* enloop1 = loop(head1);
		LinkNode* enloop1_go = enloop1;
		LinkNode* enloop2 = loop(head2);
		while(enloop1_go->next != enloop1){
			enloop1_go = enloop1_go->next;
			if(enloop1_go == enloop2){
				return enloop1;
			}
		}
		//66结构
		return nullptr;
	}
}

LinkNode* judge(LinkNode* head1, LinkNode* head2){
    if(head1 == nullptr || head2 == nullptr){
        return nullptr;
    }
	// 两个无环单链表是否相交
	if(loop(head1) == nullptr && loop(head2) == nullptr){
		return noloop(head1, head2);
	}else if(loop(head1) != nullptr && loop(head2)!= nullptr){
		return hasloop(head1, head2);
	}
	return nullptr;
}

ok，88~~