[LeetCode] 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.

 

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 Linked List

 

 

分析：

 

如果两个链表相交，那么后面的链表必然相同，找到lenA，lenB,然后跳过lenA和lenB的差，

然后比较两个指针是否相同，若相同则有交叉，同时找到交叉节点，否则就是没有交叉。

 

/**
 * 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 lenA = 0;
            int lenB = 0;

            ListNode* pA = headA;
            ListNode* pB = headB;

            while(pA)
            {   
                lenA ++; 
                pA = pA->next;
            }   

            while(pB)
            {   
                lenB ++; 
                pB = pB->next;
            }   

            if(lenA > lenB)
            {   
                for(int i = 0; i < (lenA-lenB); i++)
                    headA = headA->next;
            }   
            else if(lenA < lenB)
            {   
                for(int i = 0; i < (lenB-lenA); i++)
                    headB = headB->next;
            }

            while(headA != headB)
            {
                    headA = headA->next;
                    headB = headB->next;
            }
            if(headA == headB)
                return headA;
            else
                return NULL;
        }
};