"Original" Leetcodeoj---intersection of the Linked Lists Problem Solving report (classic intersect linked list find intersection)

Title Address:

https://oj.leetcode.com/problems/intersection-of-two-linked-lists/

Topic content:

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.

Method:

First of all, since it's time complexity O (n), then you can't try it one by one.

Second, since space complexity requires O (1), since it is not possible to use data structures such as stack or unordered_map to find the intersection of the linked list.

Well, we need a cool trick to find the intersection.

First Conversion questions:

Suppose that A and B linked lists intersect, requesting the distance from the intersection point to the node of the A-linked table head. (The so-called distance, is the head node walk several times can arrive)

First look at the specific method of seeking:

0, calculate the length of a linked list Lena

1, calculate the length of the B-linked list LenB

2. Reverse a linked list (key)

3. Recalculate the length of the B-linked list Newlenb

4. Return result = (Newlenb-lenb + lenA-1)/2

Specifically to this problem, you also need to reverse the list of a, because you can not change the original structure of the list. Then re-read the list, return to the first result node is OK.

So how did this concrete approach come about?

Try it yourself and you'll understand. In fact, it is a linked table node in the intersection next to the distribution number, drawing too much trouble, if there is time to fill one, or who do not understand the reply I will fill

All code:

/** 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; intAddressa;//fin of A.        intADDRESSB;//fin of B.        intLenA = Countlength (heada,&Addressa); intLenB = Countlength (headb,&ADDRESSB); if(Addressa! = ADDRESSB)//if has a intersect            returnNULL; ListNode*tmpheada = (ListNode *) Addressa;//To store Heada ' s tail for reverse.Reverselink (Heada); intNewlenb = Countlength (headb,&ADDRESSB); intTonew =Findcount (LENA,LENB,NEWLENB);        Reverselink (Tmpheada); returnFindnthnode (heada,tonew); }        intFindcount (intLenA,intLenB,intNewlenb) {        intGap = Newlenb-LenB; return(Gap + LenA-1) /2; } ListNode*findnthnode (ListNode *head,intToN) {        intCount =0;  while(ToN! =count) {Head= head->Next; Count++; }        returnHead; }        intCountlength (ListNode *head,int*Fin) {        intCount =0;  while(head) {*fin = (int) head; Head= head->Next; Count++; }        returncount; }        voidReverselink (ListNode *head) {ListNode*pre =NULL; ListNode*now =Head; ListNode*NXT = head->Next;  while(1) { now->next =Pre; Pre=Now ; now=NXT; if(NXT) NXT= nxt->Next; Else                 Break; }    }};

"Original" Leetcodeoj---intersection of the Linked Lists Problem Solving report (classic intersect linked list find intersection)