HappyLeetcode42: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.

Credits:
Special thanks to @stellari for adding this problem and creating all test cases.

这道题除了阅读题目的困难较大之外， 其他的都还好。非常好做的一道题目。

代码如下：

//* 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) {
        if (headA == NULL || headB == NULL)
            return NULL;
        int lengtha=1, lengthb=1;
        ListNode *cora = headA, *corb = headB;
        while (cora->next != NULL)
        {
            cora = cora->next;
            lengtha++;
        }
        while (corb->next != NULL)
        {
            corb = corb->next;
            lengthb++;
        }
        if (cora->val != corb->val)
            return NULL;
        int dif = lengtha - lengthb;
        cora = headA, corb = headB;
        if (dif >= 0)
        {
            while (dif)
            {
                dif--;
                cora = cora->next;
            }
        }
        else

        {
            while (dif)
            {
                dif++;
                corb = corb->next;
            }
        }

        while (cora != NULL)
        {
            if (cora == corb)
                return cora;
            else
            {
                cora = cora->next;
                corb = corb->next;
            }
        }
    }
};