[Leetcode] Intersection of Linked Lists

Source: Internet
Author: User

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.

Hide TagsLinked ListAnalysis: If the two linked lists Intersect, then the list of the following must be the same, find Lena,lenb, and then skip the difference between Lena and LenB, and then compare two pointers are the same, if the same cross, and find the cross node, otherwise there is no cross.
/** 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; intLenA =0; intLenB =0; ListNode* PA =Heada; ListNode* PB =headb;  while(PA) {LenA++; PA= pa->Next; }                while(PB) {LenB++; PB= pb->Next; }               if(LenA >LenB) {                    for(inti =0; I < (LENA-LENB); i++) Heada= heada->Next; }               Else if(LenA <LenB) {                    for(inti =0; I < (Lenb-lena); i++) headb= headb->Next; }             while(Heada! =headb) {Heada= heada->Next; HEADB= headb->Next; }            if(Heada = =headb)returnHeada; Else                returnNULL; }};

[Leetcode] Intersection of Linked Lists

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