Intersection of Two Linked Lists

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Write a program to find the node at which the intersection of two singly linked lists begins.

 

For example, the following two linked lists:

A:          a1 → a2                                        c1 → c2 → c3                               B:     b1 → b2 → b3

begin to intersect at node c1.

 

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.

 C++ 實現代碼：

#include<iostream>#include<new>#include<cmath>using namespace std;//Definition for singly-linked list.struct ListNode{    int val;    ListNode *next;    ListNode(int x) : val(x), next(NULL) {}};class Solution{public:    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB)    {        if(headA==NULL||headB==NULL)            return NULL;        int lenA=length(headA);        int lenB=length(headB);        int len=0;        ListNode *p=headA;        ListNode *q=headB;        if(lenA>=lenB)        {            len=lenA-lenB;            while(p&&q)            {                if(p&&len)                {                    len--;                    p=p->next;                    continue;                }                if(p->val!=q->val)                {                    p=p->next;                    q=q->next;                    continue;                }                break;            }        }        else        {            len=lenB-lenA;            while(p&&q)            {                if(q&&len)                {                    len--;                    q=q->next;                    continue;                }                if(p->val!=q->val)                {                    p=p->next;                    q=q->next;                    continue;                }                break;            }        }        if(p&&q)            return p;        else            return NULL;    }    int length(ListNode *head)    {        if(head==NULL)            return 0;        ListNode *p=head;        int len=0;        while(p)        {            len++;            p=p->next;        }        return len;    }    void createList(ListNode *&head,int *arr,int n)    {        ListNode *p=NULL;        int i=0;        for(i=0; i<n; i++)        {            if(head==NULL)            {                head=new ListNode(arr[i]);                if(head==NULL)                    return;            }            else            {                p=new ListNode(arr[i]);                p->next=head;                head=p;            }        }    }};int main(){    Solution s;    ListNode *L1=NULL;    ListNode *L2=NULL;    ListNode *L=NULL;    int arr1[10]= {12,11,9,7,5,3,1,0};    int arr2[10]= {12,11,8,6,4,2};    s.createList(L1,arr1,8);    s.createList(L2,arr2,6);    L=s.getIntersectionNode(L1,L2);    cout<<L->val<<endl;}

 

Intersection of Two Linked Lists