Intersection of Two Linked Lists

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++ 实现代码：

#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;

}