【leetcode】Intersection of Two Linked Lists

 

Intersection of Two Linked Lists

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.

 

先从头到尾扫描，看两个链表长度相差多少。

然后让长的链表先走过这个长度差，然后两个链表同时走，直到相遇

 

 1 /**

 2  * Definition for singly-linked list.

 3  * struct ListNode {

 4  *     int val;

 5  *     ListNode *next;

 6  *     ListNode(int x) : val(x), next(NULL) {}

 7  * };

 8  */

 9 class Solution {

10 public:

11     ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {

12        

13        

14         if(headA==NULL||headB==NULL)

15         {

16             return NULL;

17         }

18        

19         ListNode *hA=headA,*hB=headB;

20        

21         int lenA=1,lenB=1;

22         while(hA->next!=NULL)

23         {

24             lenA++;

25             hA=hA->next;

26         }

27        

28         while(hB->next!=NULL)

29         {

30             lenB++;

31             hB=hB->next;

32         }

33        

34        

35         if(hA!=hB)

36         {

37             return NULL;

38         }

39        

40         int dis=lenA-lenB;

41        

42        

43         hA=headA;

44         hB=headB;

45        

46         if(dis>0)

47         {

48             while(dis)

49             {

50                 hA=hA->next;

51                 dis--;

52             }

53         }

54        

55         if(dis<0)

56         {

57             dis=-dis;

58             while(dis)

59             {

60                 hB=hB->next;

61                 dis--;

62             }

63         }         

64        

65         while(hA!=hB)

66         {

67             hA=hA->next;

68             hB=hB->next;

69         }

70        

71        

72         return hA;

73     }

74 };

75