Leetcode intersection of the 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.

Credits:
Special thanks to @stellari for adding this problem and creating all test cases.

It's easy if you don't ask for this question. You can do it with hashset.

But the topic requires in O (n) time and use only O (1) memory.

You can traverse two linked lists to make their lengths as long as possible. And then compare them one by one.

1 /**2 * Definition for singly-linked list.3 * public class ListNode {4 * int val;5 * ListNode Next;6 * ListNode (int x) {7 * val = x;8 * next = null;9  *     }Ten  * } One  */ A  Public classSolution { -      PublicListNode Getintersectionnode (ListNode heada, ListNode headb) { -         inta=0, b = 0; theListNode Tempa =Heada; -ListNode TEMPB =headb; -          while(Tempa! =NULL) { -++A; +Tempa =Tempa.next; -         } +          while(TEMPB! =NULL) { A++b; atTEMPB =Tempb.next; -         } -Tempa =Heada; -TEMPB =headb; -          while(A >b) { ---A; inTempa =Tempa.next; -         } to          while(B >a) { +--b; -TEMPB =Tempb.next; the         } *          while(Tempa! =NULL) { $             if(Tempa = =TEMPB) {Panax Notoginseng                 returnTempa; -}Else { theTempa =Tempa.next; +TEMPB =Tempb.next; A             } the         } +         return NULL; -     } $}

Leetcode intersection of the Linked Lists