Leetcode Intersection of Two Linked Lists

/**

 * Definition for singly-linked list.

 * struct ListNode {

 *     int val;

 *     ListNode *next;

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

 * };

 */

class Solution {

public:

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

        int lena = count(headA);

        int lenb = count(headB);

        int diff = 0;

        ListNode* longer = NULL;

        ListNode* shorter= NULL;

        if (lena > lenb) {

            diff = lena - lenb;

            longer = headA;

            shorter= headB;

        } else {

            diff = lenb - lena;

            longer = headB;

            shorter= headA;

        }

        longer = forward(longer, diff);

        

        while (longer != shorter) {

            longer = forward(longer, 1);

            shorter= forward(shorter, 1);

        }

        return longer;

    }

    ListNode* forward(ListNode* head, int step) {

        while(head != NULL) {

            if (step-- <= 0) {

                break;

            }

            head = head->next;

        }

        return head;

    }

    int count(ListNode* head) {

        int res = 0;

        while (head != NULL) {

            head = head->next;

            res++;

        }

        return res;

    }

};

Leetcodeも可視化され、ますますハイエンドになりました
第2ラウンド:
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.

 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  // 20:08

10 class Solution {

11 public:

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

13         ListNode* cura = headA;

14         ListNode* curb = headB;

15         

16         int na = 0, nb = 0;

17         

18         while (cura != NULL) {

19             na++;

20             cura = cura->next;

21         }

22         

23         while (curb != NULL) {

24             nb++;

25             curb = curb->next;

26         }

27         ListNode* prewalk = NULL;

28         

29         int diff = 0;

30         if (na > nb) {

31             prewalk = headA;

32             diff = na - nb;    

33         } else if (na < nb) {

34             prewalk = headB;

35             diff = nb - na;

36         }

37         while (diff--) {

38             prewalk = prewalk->next;

39         }

40         

41         cura = headA;

42         curb = headB;

43         

44         if (na > nb) {

45             cura = prewalk;

46         } else if (na < nb){

47             curb = prewalk;

48         }

49         

50         while (cura != curb) {

51             cura = cura->next;

52             curb = curb->next;

53         }

54         return cura;

55     }

56 };