【The-Art-Of-Programming】判断两链表是否相交
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题目：
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。为了简化问题，我们假设俩个链表均不带环。
问题扩展：
- 如果链表可能有环列?
- 如果需要求出俩个链表相交的第一个节点列?
解析：
流程：
1. 判断两链表是否有环
2. 若链表都不成环，判断尾结点是否相等
3.若链表都成环，判断一条链表上俩指针相遇的结点在不在另一天链表的环上
4.若一条链表成环，另一条不成环，则两链表不相交
#include <iostream>using namespace std;struct node{int data;struct node * next;};typedef struct node Node;//1.判断链表是否带环，若带环，返回环里的结点 NodeList *testCircle(NodeList *head){Node *slow, *fast;if(head == NULL){return NULL;}slow = head;fast = head;while(slow != NULL && fast != NULL){slow = slow->next;fast = fast->next->next;if(slow == fast){return slow;}}}//判断一条链表上俩指针相遇的结点在不在另一天链表的环上int isJoinedSimple(Node *head1, Node *head2){if(head1 == NULL || head2 == NULL){return 0;}while(head1->next != NULL){head1 = head1->next;}while(head2->next != NULL){head2 = head2->next; } return head1 == head2;}//判断链表是否相交//若链表都不成环，跳到isJoinedSimple（），判断尾结点是否相等//若链表都成环，判断一条链表上俩指针相遇的结点在不在另一天链表的环上//若一条链表成环，另一条不成环，则两链表不相交int isJoined(Node *head1, Node *head2){Node *circle1, *circle2;circle1 = testCircle(head1);circle2 = testCircle(head2);if(circle1 == NULL && circle2 == NULL){return isJoinedSimple(head1, head2); }if(circle1 != NULL && circle2 != NULL){Node * p = circle1->next;while(p != circle1){if(p == circle2){return 1;}p = p->next;} } return 0； }
扩展1. 求存在环的链表的环入口点NodeList *testCircle(NodeList *head){Node *slow, *fast;if(head == NULL){return NULL;}slow = head;fast = head;while(slow != NULL && fast != NULL){slow = slow->next;fast = fast->next->next;if(slow == fast){break;}}if(slow == NULL || fast->next == NULL){ return NULL;}slow = head;while(slow != fast){slow = slow->next;fast= fast->next;} return slow;}
扩展2. 求两个无环链表的相交的第一个节点1.先遍历两条链表，求出链表长度，得到长度差K；
2. 长链表先遍历k个结点，再和短链表一起遍历，两个相等的第一个结点就是所求的。
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