Intersection of Linked Lists

First, the topic

1, examining

　　

2. Analysis

Give two one-way linked list, if the two linked list with overlapping parts, output the first node of overlapping nodes, otherwise output null;

Second, the answer

1, Ideas:

Method One,

Aligns the two linked lists at the tail end, and begins traversing the linked list to find out if there are overlapping nodes.

①, calculate the length of two linked lists;

②, the long list is moved backwards, so that the tail of the two linked lists are aligned;

③, start the lookup node.

     PublicListNode Getintersectionnode (ListNode heada, ListNode headb) {intLenA = 0, LenB = 0; ListNode node=Heada; //step1: Calculating the length of A and B nodes         while(Node! =NULL) {node=Node.next; LenA++; } node=headb;  while(Node! =NULL) {node=Node.next; LenB++; }                //step2: Causes A, B to be aligned at the tail, and to begin to traverse backwards        if(LenA >LenB) {Node=Heada; intGap = LenA-LenB;  while(gap--> 0) Heada=Heada.next; }        Else if(LenA <LenB) {Node=headb; intGap = LenB-LenA;  while(gap--> 0) headb=Headb.next; }                //step3: Finding the target node         while(Heada! =NULL&& headb! =NULL) {            if(Heada = =headb)returnHeada; Heada=Heada.next; HEADB=Headb.next; }        returnHeada; }

Method Two,

There is no need to calculate the chain table length.

①, use pointer A to point to Heada, pointer to headb;

②, Adopt A while loop, when a! = B, both A and B move backwards, and when a = null, a points to headb, and when b = null, B points to Heada.

Eventually, a will traverse the Heada + headb,b to traverse the HEADB + Heada, eventually a, b encounters, the encounter point is null (no overlapping nodes), or node (the first overlapping node).

     PublicListNode GetIntersectionNode2 (ListNode heada, ListNode headb) {if(Heada = =NULL|| HEADB = =NULL)            return NULL; ListNode a= Heada, B =headb;  while(A! =b) {a= (A = =NULL?headb:a.next); b= (b = =NULL?heada:b.next); }        returnA; }

Intersection of Linked Lists