Leetcode 160:intersection of the Linked Lists

Intersection of Linked ListsTotal Accepted: 8676 Total Submissions: 32571

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.

Analysis

Solution One:
First cycle, find out the length difference of two lists n
The second cycle, the long list go first n steps, and then move at the same time to determine whether there is the same node

Solution Two:

After the list to the tail, jump to another list of the head, the meeting point is intersection points.

[Precautions]
NONE

[CODE]

Solution One:

/** * Definition for singly-linked list. * public class ListNode {* int val; * List Node Next;  * ListNode (int x) {* val = x; * next = NULL; *} *} */public class Solution {public ListNode                Getintersectionnode (ListNode Heada, ListNode headb) {if (Heada==null | | headb==null) return NULL;        ListNode p = Heada;        ListNode q = headb;        int pcount = 0;        int qcount = 0;            while (P.next! = NULL | | Q.next! = NULL) {if (p = = q) return p; if (p.next! = null) p = p.next;            else ++qcount; if (q.next! = null) q = q.next;        else ++pcount;        if (P! = q) return null;        p = Heada;        Q = headb;        while (pcount--! = 0) {p = p.next;        } while (Qcount--! = 0) {q = Q.next;            } while (P! = q) {p = p.next;        Q = q.next;    } return p; }}

Solution Two:

/** * Definition for singly-linked list. * public class ListNode {*     int val; *     ListNode Next; *     listnode (int x) {*         val = x; *         next = null; *< c5/>} *} */public class Solution {public    ListNode getintersectionnode (ListNode heada, ListNode headb) {        if (Heada ==null | | Headb==null) return null;                ListNode p = heada;        ListNode q = headb;        if (p = = q) return p;                while (P!=null && q!=null) {            p = p.next;            Q = q.next;        }                if (p==null) p = headb; else q = Heada;                while (P!=null && q!=null) {            p = p.next;            Q = q.next;        }        if (p==null) p = headb; else q = Heada;        while (P!=null && q!=null) {            if (p==q) return p;            p = p.next;            Q = q.next;        }        return null;}    }

Leetcode 160:intersection of Linked Lists