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, return null.
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 should preferably run in O(n) time and use only O(1) memory.
題目大意：查找2個鏈表相交的第一個交點
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) //計算2個鏈表的長度
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) //長的先走|a-b|步，讓2個鏈表的剩下的長度一樣
31 {
32 i = a-b;
33 while(i)
34 {
35 acur = acur->next;
36 i--;
37 }
38 }
39 else
40 {
41 i = b - a;
42 while(i)
43 {
44 bcur = bcur->next;
45 i--;
46 }
47 }
48 while(acur != NULL) //判斷2個鏈表的第一個相交的結點
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 }
