"Leetcode" 160. Intersection of Linked Lists

Title:

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.

Tips:

Assuming that the two linked lists are combined, the length of the merged part must be the same, and the length will be different before the merge, so the length is first calculated and then the long list is moved forward so that they are "at the same starting line" and then compared in turn.

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 | |!headb)returnNULL; intLenA =0, LenB =0; ListNode*nodea, *NodeB;  for(NodeA = Heada; nodeA; NodeA = nodea->next, + +LenA);  for(NodeB = headb; nodeB; NodeB = nodeb->next, + +LenB); if(LenB >LenA) {             for(inti =0; i < Lenb-lena; ++i, headb = headb->next); } Else {             for(inti =0; i < Lena-lenb; ++i, Heada = heada->next); }        if(Heada = = headb)returnHeada;  while(Heada &&headb) {Heada= heada->Next; HEADB= headb->Next; if(Heada = = headb)returnHeada; }        returnNULL; }};

"Leetcode" 160. Intersection of Linked Lists