判断两个链表是否相交并求交点

链表是否相交分为以下三种情况：

1.两个单链表都不带环

2.只有一个单链表带环

3.两个单链表都带环

#pragma once#include #include struct ListNode{int value;ListNode* next;};class CreateList{typedef ListNode node;public:CreateList():root(NULL){}void push(int data){if (NULL == root){node *temp = new node();temp->next =NULL;temp->value = data;root = temp;}else{node* temp = new node();node* tmp = root;while (tmp->next){tmp = tmp->next;}tmp->next = temp;temp->value = data;temp->next = NULL;}}void PrintList(){node* tem = root;while (tem){std::cout<value<<"->";tem = tem->next;}std::cout<<"NULL";}node* pHead(){return root;}int ListSize(){if (NULL == root){return 0;}int count = 0;node* temp = root;while (NULL != temp){count++;temp = temp->next;}return count;}private:node* root;};bool JudgeIntersect(ListNode* n1, ListNode* n2)//判断不带环相交问题{if (NULL == n1){return false;}if (NULL == n2){return false;}ListNode* t1 = n1;ListNode* t2 = n2;while(t1->next){t1 = t1->next;}while (t2->next){t2 = t2->next;}if (t2->value == t1->value){return true;}return false;}//求相交点：//1:拿一个链表中的每个节点和另一个链表中的所有节点作比较//2:求出两个链表大小：让大的链表先走n步，然后一起走，每次比较是否相同ListNode* FindIntersectNode(ListNode* n1, int s1, ListNode* n2, int s2){if ( NULL == n1 || NULL == n2){return NULL;}int mid = 0;ListNode* tmp1 = n1;ListNode* tmp2 = n2;if (s1 != s2){mid = s2 - s1;if (mid < 0){mid = 0 - mid;while (mid != 0){tmp1 = tmp1->next;mid--;}}else{while (mid != 0){tmp2 = tmp2->next;mid--;}}}while (tmp1->next){tmp1 = tmp1->next;tmp2 = tmp2->next;if (tmp1->value == tmp2->value){return tmp1;}}return NULL;}//判断带环相交//1.两个都带环，且在环外相交//2.两个都带环，且在环内相交ListNode* GetInNode(ListNode* p1){if (NULL == p1 || NULL == p1->next){return NULL;}ListNode* lf = p1;ListNode* ls = p1;while (NULL != ls){lf = lf->next;ls = ls->next->next;if (lf->value == ls->value){break;}}ListNode* ph = p1;while (lf!= ph){lf = lf->next;ph = ph->next;}return ph;}ListNode* GetIntersectNode(ListNode* plist1, ListNode* plist2){if (NULL == plist1 || NULL == plist2){return NULL;}ListNode* pn1 = NULL;ListNode* pn2 = NULL;pn1 = GetInNode(plist1);pn2 = GetInNode(pn1);if (pn1 == pn2){//环内相交//整个环都是交点return pn1;}else if(pn1 != pn2){//环外相交int count1 = 0;int count2 = 0;ListNode* s1 = plist1;while (s1 != pn1){count1++;s1 = s1->next;}ListNode* s2 = plist2;while (s2 != pn1){count2++;s2 = s2->next;}return FindIntersectNode(plist1, count1, plist2, count2);}}#include "x.h"int main(){CreateList L;int ls;L.push(1);L.push(2);L.push(3);L.push(4);L.push(5);L.push(6);L.PrintList();ls = L.ListSize();std::cout<value<