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.

Credits:
Special thanks to @stellari for adding this problem and creating all test cases.

可以将A,B两个链表看做两部分，交叉前与交叉后。

交叉后的长度是一样的，因此交叉前的长度差即为总长度差。

只要去除这些长度差，距离交叉点就等距了。

为了节省计算，在计算链表长度的时候，顺便比较一下两个链表的尾节点是否一样，

若不一样，则不可能相交，直接可以返回NULL

 就是长的链表开始多走 （h1的数量 - h2的数量）步，然后和短链表同步往下走，遇到的第一个相同的节点就是最早的公共节点

/** * 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;        ListNode* t1=headA;        ListNode* t2=headB;        int len1=1,len2=1;        while(t1->next!=NULL)        {            t1=t1->next;            len1++;        }        while(t2->next!=NULL)        {            t2=t2->next;            len2++;        }               if(len1>len2)        {            for(int i=0;i<(len1-len2);i++)            {                headA=headA->next;            }        }       else if(len2>len1)        {            for(int i=0;i<(len2-len1);i++)            {                headB=headB->next;            }        }        while(headA!=headB)        {            headA=headA->next;             headB=headB->next;        }        return headA;            }};