Leetcode 160. Intersection of Linked Lists

This topic is easier, my idea is to compare the length of the two list, we have to find the intersection is not possible for the long chain of the front of the long part of the list,

So to compare the shorter part of the two

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

  /** * 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) {size_t lengtha=0, lengthb=0,minus=0; ListNode *pa=heada, *pb=headb,*intersectionnode=NULL; while (PA! =NULL) {+ +Lengtha; PA = pa->Next } while (PB! =NULL) {+ +LENGTHB; PB = pb->Next } if (Lengtha >LENGTHB) {PA =Heada; PB =HEADB; minus = Lengtha- LENGTHB; minus--) {PA = Pa->  next;} while (PA! = NULL&&PB! =  NULL) {if (PA-&G T;val = = pb->val&&intersectionnode==  NULL) Intersectionnode =  PA; if (pa->val! = Pb->  val) Intersectionnode =  NULL; pA = Pa->  Next; PB = Pb->  next;} return  Intersectionnode;} els E  {PA =  heada; pb =  headb; minus = LENGTHB- lengtha; while (minus--) {PB = Pb->  n Ext } while (PA! = NULL&&PB! =  null) {if (Pa->val = = Pb->val&&intersectionnode = =  NULL) Int Ersectionnode =  pa; if (pa->val! = Pb->  val) intersectionnode =  NULL; pA = Pa->  Next; PB = Pb->  Next;} return  Intersectionnode;}}};          

Leetcode 160. Intersection of Linked Lists