Intersection of Two Linked Lists

Description:

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.

Code1:

  ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
          //算法1：假设链表无环，对齐A,B链表
          ListNode * pa = headA;
          ListNode * pb = headB;
          int lengthA = 0, lengthB = 0;
          while ( pa )
          {
              pa = pa->next;
              lengthA++;
         }
         while ( pb )
         {
             pb = pb->next;
             lengthB++;
         }
         
         ListNode * pLong = NULL;
         ListNode * pShort = NULL;
         pLong = (lengthA >= lengthB)?headA:headB;
         pShort = (lengthA < lengthB)?headA:headB;
         //对齐两个链表
         for ( int i = 0; i < abs(lengthA-lengthB); i++ )
         {
             pLong = pLong->next;
         }
         
         while (pShort!=pLong && pShort!=NULL && pLong != NULL)
         {
                 pShort = pShort->next;
                 pLong = pLong->next;
             }
         }
         return pShort;
     }

 

 Code2:利用栈实现逆向访问链表

 ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
       stack<ListNode*>a;
       stack<ListNode*>b;
       ListNode* p = headA;
       ListNode* q = headB;
       while (p)
       {
           a.push(p);
           p = p->next;
       }
       while (q)
       {
            b.push(q);
            q = q->next;
       }
       ListNode* lastNode = NULL;
       while ( !a.empty() && !b.empty() )
       {
           if (a.top() != b.top())
           {
               return lastNode;
           }
           else
           {
               lastNode = a.top();
               a.pop();
               b.pop();
           }
       }
       return lastNode;
    }