"Leetcode" intersection of the Linked Lists

Intersection of Linked Lists

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

For example, the following, linked lists:

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

Begin to intersect at node C1.

Notes:

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

A A, b two linked list can be seen as two parts, before and after crossing.

The length of the crossover is the same, so the length difference before the intersection is the total length difference.

As long as these lengths are removed, the distance to the intersection is equidistant.

In order to save the calculation, in the calculation of the chain table length, the way to compare the two linked list of the tail node is the same,

If not, it is not possible to intersect and can return null directly

/** Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * ListNode (int x) : Val (x), Next (NULL) {}}; */classSolution { Public: ListNode*getintersectionnode (ListNode *heada, ListNode *headb) {        if(Heada = = NULL | | headb = =NULL)returnNULL; intLenA =1; ListNode* CurA =Heada;  while(Cura->next! =NULL) {LenA++; CurA= cura->Next; }        //Now CurA is the tail of A                intLenB =1; ListNode* CurB =headb;  while(Curb->next! =NULL) {LenB++; CurB= curb->Next; }        //Now CurB is the tail of B                if(CurA! = CurB)//No intersection            returnNULL; Else        {            //Align            intdiff = lena-LenB; if(diff >0)            {//A Go                 while(diff) {Heada= heada->Next; Diff--; }            }            Else            {//B Go                 while(diff) {headb= headb->Next; Diff++; }            }            //Go together             while(true)            {//A and B have judged to intersect, no need to avoid NULL                if(Heada = =headb)returnHeada; Else{Heada= heada->Next; HEADB= headb->Next; }            }        }    }};

"Leetcode" intersection of the Linked Lists