特殊单链表的一些操作

1，建一个有环单链表：

#include<iostream>

using namespace std;

#include<stack>

struct Node

{

int _data;

struct Node *next;

};

typedef Node *PList;

Node* CreatNode(int x)//创建节点

{

Node *tmp = new Node;

tmp->_data = x;

tmp->next = NULL;

return tmp;

}

void Pushback(PList &head, int x)//尾插

{

if (head == NULL)

{

head = CreatNode(x);

}

else

{

Node * tail = head;

while (tail->next != NULL)

{

tail = tail->next;

}

tail->next = CreatNode(x);

}

}

Node* Find(PList head, int x)

{

Node *cur = head;

while (head != NULL)

{

if (cur->_data == x)

{

return cur;

}

cur = cur->next;

}

}void main()

{

PList list1;

list1= NULL;

Node *cur = NULL;

Node *cur1 = NULL;

Pushback(list1, 1);

Pushback(list1, 2);

Pushback(list1, 3);

Pushback(list1, 4);

Pushback(list1, 5);

Pushback(list1, 6);

Pushback(list1, 7);

Pushback(list1, 8);

cur=Find(list1, 4);//找到两个节点，把他们的next更改。

cur1=Find(list1, 8);

cur1->next = cur;

}

2，检查是否有环，有环输出环的节点个数：

int CheackCyce(PList head)

{

Node *fast = head;

Node *slow = head;

while (fast&&fast->next)

{

fast = fast->next->next;

slow = slow->next;

if (fast == slow)

{

int count = 1;

Node *cur = slow;

while (cur->next != slow)

{

++count;

cur = cur->next;

}

return count;

}

}

return -1;

}

3，找到环的入口点：

Node *FindEntry(PList head)//找到环的入口点

{

Node *fast = head;

Node *slow = head;

while (fast&&fast->next)

{

fast = fast->next->next;

slow = slow->next;

if (fast == slow)

{

fast = head;

while (fast != slow)

{

fast = fast->next;

slow = slow->next;

}

return fast;

}

}

}//求入口点，第一次的相遇节点为meetNode//fast=k+l+x slow=k+x;k+x=L;注意环很小时候

快指针不止走了一圈，

//fast=k+nl+x,slow=k+x;k+nl+x=2(k+x),nl=k+x;(K位起点到环起点的距离。x为环的起点到相遇点，l为环的长度) k=nl-x;

//

4,判断两个单链表是否相交，

int Iscomp(PList head1, PList head2)//（无环）是否相交

{

Node *cur1 = head1;

Node *cur2 = head2;

if (head1 == NULL)

{

return 0;

}

if (head2 == NULL)

{

return 0;

}

while (cur1&&cur1->next)

{

cur1 = cur1->next;

}

while (cur2&&cur2->next)

{

cur2 = cur2->next;

}

while (cur1 ==cur2)

{

return 1;

}

return -1;

}

5，两个单链表相交的起点：

Node * copx(PList head1, PList head2)//（无环）是否相交

{

int count1 = 0;

int count2 = 0;

int sum = 0;

Node *cur1 = head1;

Node *cur2 = head2;

Node *cur3 = head1;

Node *cur4 = head2;

if (head1 == NULL)

{

return 0;

}

if (head2 == NULL)

{

return 0;

}

while (cur1&&cur1->next)//计算链表1的长度

{

cur1 = cur1->next;

count1++;

}

while (cur2&&cur2->next)//计算链表2的长度

{

cur2 = cur2->next;

count2++;

}

if (count1 > count2)//求出链表1比链表2长多少

{

sum = count1 - count2;

while (sum--)//让链表一先走比链表2长度的距离

{

cur3 = cur3->next;

}

}

else

{

sum = count2 - count1;

while (sum--)

{

cur4 = cur4->next;

}

}

while (cur3 != cur4)//距离相同，所以同时到达入口点；

{

cur3 = cur3->next;

cur4 = cur4->next;

}

return cur3;

}