leetcode——160——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.

/** * 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 = 0;        int lengthB = 0;                ListNode* NodeA = headA;        ListNode* NodeB = headB;                while(NodeA!=NULL)        {            NodeA = NodeA->next;            lengthA++;        }        while(NodeB!=NULL)        {             NodeB = NodeB->next;            lengthB++;        }                NodeA = headA;        NodeB = headB;                if(lengthA>lengthB)        {            for(int i=0;i<(lengthA-lengthB);i++)            {                NodeA = NodeA->next;            }        }        if(lengthA<lengthB)        {           for(int i=0;i<(lengthB-lengthA);i++)            {                NodeB = NodeB->next;            }        }        while(NodeA->val!=NodeB->val)        {            NodeA = NodeA->next;            NodeB = NodeB->next;            if(NodeA ==NULL|| NodeB == NULL)               return NULL;                                }                return NodeA;                    }};