第7题 微软亚院之编程判断俩个链表是否相交 给出俩个单向链表的头指针，比如h1，h2，判断这俩个链表是否相交，并找出第一个相交结点。

思路

1.判断两链表是否相交

可以用两个指针，p1End, p2End，分别指向链表h1, h2的最后一个结点。如果p1End = p2End说明两链表相交。

2.找出相交的第一个结点

可以先求得链表h1, h2的长度之差的绝对值sub, 将较长的那个链表向后移动sub个位置后。从两链表的当前位置开始一直往后比较，第一个相等的结点即是第一个相交结点。

#include <iostream>
using namespace std;

typedef struct tagNode
{
tagNode *pNext;
int len;
int data;
}Node;

//判断n1, n2是否相交,相交返回true，否则返回false
bool isIntersect(Node *n1, Node *n2)
{
Node *p1End = NULL;
Node *p2End = NULL;

//将p1End指向n1的最后一个节点
while (NULL != n1)
{
p1End = n1;
n1 = n1->pNext;
}

//将p2End指向n2的最后一个节点
while (NULL != n2)
{
p2End = n2;
n2 = n2->pNext;
}

//相等说明n1, n2相交
if (p1End == p2End)
{
return true;
}
else
{
return false;
}
}

//返回n1, n2相交的第一个节点
Node *FirstIntersect(Node *n1, Node *n2)
{
int n1Len = n1->len;
int n2Len = n2->len;
int sub = 0;

//把长度较长的链表往后移动sub个位置,使n1,n2相对链表尾的节点个数相同
if (n1Len > n2Len)
{
sub = n1Len - n2Len;
while (sub--)
{
n1 = n1->pNext;
}
}
else
{
sub = n2Len - n1Len;
while (sub--)
{
n2 = n2->pNext;
}
}

//n1,n2第一次相等的那个结点,就是第一个相交的结点
while (n1 != n2)
{
n1 = n1->pNext;
n2 = n2->pNext;
}

return n1;
}

//为了方便实验,此函数的功能是让两个链表相交
void MakeIntersect(Node *n1, Node *n2)
{
int i = 2;
int j = 3;

while (i--)
{
n1 = n1->pNext;
}

while (j--)
{
n2 = n2->pNext;
}

n1->pNext = n2;

}

//创建链表,并返回头指针
Node *CreateList()
{
Node *head = NULL;
Node *tail = NULL;
Node *temp = NULL;
static int k = 1;

tail = head = new Node;
head->data = k;
head->len = 1;
head->pNext = NULL;

for (int i = 0; i < 6; ++i)
{
++k;
temp = new Node;
temp->data = k;
++(temp->len);
temp->pNext = NULL;

tail->pNext = temp;
tail = temp;

}

return head;
}

//遍历链表
void Tranverse(Node *head)
{
Node *pCur = head;
while (NULL != pCur)
{
cout<<pCur->data<<endl;
pCur = pCur->pNext;
}
}

int main()
{
Node *pFirstList = CreateList();
Node *pSecondList = CreateList();
Node *IntersectNode = NULL;
cout<<"分别遍历2个链表"<<endl;
Tranverse(pFirstList);
cout<<"\n链表2"<<endl;
Tranverse(pSecondList);

cout<<"\n两链表相交后"<<endl;
MakeIntersect(pFirstList, pSecondList);
Tranverse(pFirstList);
cout<<"\n链表2"<<endl;
Tranverse(pSecondList);

//返回相交的第一个结点并输出
if (isIntersect(pFirstList, pSecondList))
{
cout<<"相交结点为"<<endl;
IntersectNode = FirstIntersect(pFirstList, pSecondList);
cout<<IntersectNode->data<<endl;
}
else
{
cout<<"不相交"<<endl;
}

return 1;
}