1. 程式人生 > >判斷兩個連結串列是否相交,若相交,求交點。(連結串列可能帶環)

判斷兩個連結串列是否相交,若相交,求交點。(連結串列可能帶環)

int IsListCrossWithCircle(pList plist1, pList plist2)
{
	pNode pMeetNode1 = NULL, pMeetNode2 = NULL;
	//杜絕空連結串列的存在
	if (NULL == plist1 && NULL == plist2)
	{
		return 0;
	}
	//先各自判斷兩個連結串列是否帶環(帶環返回相遇點,否則返回空)
	pMeetNode1 = IsCircle(plist1);
	pMeetNode2 = IsCircle(plist2);
	//兩個連結串列都不帶環
	if (NULL == pMeetNode1 && NULL == pMeetNode2)
	{
		//找兩個連結串列的最後一個結點,然後檢測是否是同一個結點
		pNode pTail1 = plist1;
		pNode pTail2 = plist2;
		if (pTail1 == NULL || pTail2 == NULL)
		{
			return 0;
		}
		while (pTail1)
		{
			pTail1 = pTail1->next;
		}
		while (pTail2)
		{
			pTail2 = pTail2->next;
		}
		if (pTail1 == pTail2)
		{
			return 1;
		}
	}
	//兩個連結串列均帶環
	else if (pMeetNode1 && pMeetNode2)
	{
		pNode pCur = pMeetNode1;
		do
		{
			if (pCur == pMeetNode2)
			{
				return 2;
			}
			pCur = pCur->next;
		}while (pCur->next != pMeetNode1);//繞環一週
	}
	return 0;
}