题目：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. 

题目解答:

　　判断两个链表是否有交集，并返回相交的第一个节点，若不想交，返回NULL。

　　首先计算两个链表的长度，让长链表走到短链表头平行的位置之后，两个链表再一起开始走。并一边走，一边来判断是否节点是否相同。

代码如下：

/**

 * 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) {

        int lenA = getListLen(headA);

        int lenB = getListLen(headB);

        if((lenA == 0) || (lenB == 0))

            return NULL;

        ListNode *p = headA;

        ListNode *q = headB;

        if(lenA > lenB)

        {

            int i = lenA - lenB;

            while(i != 0)

            {

                i--;

                p = p -> next;

            }

        }

        else

        {

            int i = lenB - lenA;

            while(i != 0)

            {

                i--;

                q = q -> next;

            }

        }

        while((p != NULL) && (q != NULL) && (p != q) )

        {

            p = p -> next;

            q = q -> next;

        }

        if(p == q)

            return p;

        else

            return NULL;

    }

   

    int getListLen(ListNode *head)

    {

        int len = 0;

        ListNode *p = head;

        while(p != NULL)

        {

            len++;

            p = p-> next;

        }

        return len;

    }

};