Intersection of Two Linked Lists

題目：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.
  示例代碼如下：

  class Solution {
  public:
      ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
  		if(headA==NULL ||headB==NULL)
  			return NULL;
  		int lenHeadA=0,lenHeadB=0;
  		ListNode *p=headA,*q=headB;
  		//求鏈表headA和headB的長度
  		while(p){
  			p=p->next;
  			lenHeadA++;
  		}
  		p=headA;
  		while(q){
  			q=q->next;
  			lenHeadB++;
  		}
  		q=headB;
  		int distance=0;
  		if(lenHeadA>lenHeadB){
  			distance=lenHeadA-lenHeadB;
  			for(int i=0;inext;
  		}
  		if(lenHeadAnext;
  		}
  		//循環結束的條件是p==q!=null(有相交點) 
  		//或者是p==q==null(無相交點)
  		while(p!=q){
  			p=p->next;
  			q=q->next;
  		}
  		return p;
  };