160. Intersection of Two Linked Lists, intersectionlinked

160. Intersection of Two Linked Lists, intersectionlinked

Write a program to find the node at which the intersection of two singly linked lists begins.

 

For example, the following two linked lists:

A:          a1 → a2                                        c1 → c2 → c3                               B:     b1 → b2 → b3

Begin to intersect at node c1.

 

Notes:

If the two linked lists have no intersection at all, returnnull.

The linked lists must retain their original structure after the function returns.

You may assume there are no cycles anywhere in the entire linked structure.

Your code shocould preferably run in O (n) time and use only O (1) memory.

Search for the first intersection of two linked lists

 

1/** 2 * Definition for singly-linked list. 3 * struct ListNode {4 * int val; 5 * struct ListNode * next; 6 *}; 7 */8 struct ListNode * getIntersectionNode (struct ListNode * headA, struct ListNode * headB) {9 int a, B, I; 10 struct ListNode * acur; 11 struct ListNode * bcur; 12 acur = headA; 13 bcur = headB; 14 a = B = 0; 15 while (acur! = NULL | bcur! = NULL) // calculate the length of two linked lists: 16 {17 if (acur! = NULL) 18 {19 a ++; 20 acur = acur-> next; 21} 22 if (bcur! = NULL) 23 {24 B ++; 25 bcur = bcur-> next; 26} 27} 28 acur = headA; 29 bcur = headB; 30 if (a> B) // long first | a-B | step, let the remaining length of the two linked lists be the same as 31 {32 I = a-B; 33 while (I) 34 {35 acur = acur-> next; 36 I --; 37} 38} 39 else40 {41 I = B-a; 42 while (I) 43 {44 bcur = bcur-> next; 45 I --; 46} 47} 48 while (acur! = NULL) // judge the first intersection node of the two linked lists 49 {50 if (acur = bcur) 51 break; 52 acur = acur-> next; 53 bcur = bcur-> next; 54} 55 if (acur! = NULL) 56 return acur; 57 return NULL; 58 59}