【剑指】23.链表中环的入口结点

题目描述

• 一个链表中包含环，找出环的入口结点。

算法分析

• 设链表为1,2,3,...,n,...m 其中m指向n，形成环；可知链表长度为m，环长度为m-n+1，环的入口为n；
• 使得快慢指针均指向头结点，慢指针步长为1，快指针步长为2；两者相遇时，慢指针走了x步，此时块慢指针均指向1+x，即快指针比慢指针多走了一个环的长度，即1+x等于1+2x坐标，所以x等于环的长度m-n+1；
• 那么，此时快慢指针的位置为1+x=m-n+2，慢指针走到尾节点还需要n-2步，走到环入口，+1，还需要n-1步，若此时将某个节点从头结点走n-1步，则两者同时到达环入口节点。
  

提交代码：

class Solution {
public:
	ListNode* EntryNodeOfLoop(ListNode* pHead)
	{
		if (!pHead)
			return nullptr;

		ListNode* fastNode = pHead;
		ListNode* slowNode = pHead;

		while (fastNode && fastNode->next)
		{
			fastNode = fastNode->next;
			fastNode = fastNode->next;

			slowNode = slowNode->next;

			if (fastNode == slowNode)
				break;
		}

		if (!fastNode || !fastNode->next)
			return nullptr;

		fastNode = pHead;
		while (fastNode != slowNode)
		{
			fastNode = fastNode->next;
			slowNode = slowNode->next;
		}

		return fastNode;
	}
};

测试代码：

// ==================== Test Code ====================
void Test(char* testName, ListNode* pHead, ListNode* entryNode)
{
	if (testName != nullptr)
		printf("%s begins: ", testName);
	Solution s;
	if (s.EntryNodeOfLoop(pHead) == entryNode)
		printf("Passed.\n");
	else
		printf("FAILED.\n");
}

// A list has a node, without a loop
void Test1()
{
	ListNode* pNode1 = CreateListNode(1);

	Test("Test1", pNode1, nullptr);

	DestroyList(pNode1);
}

// A list has a node, with a loop
void Test2()
{
	ListNode* pNode1 = CreateListNode(1);
	ConnectListNodes(pNode1, pNode1);

	Test("Test2", pNode1, pNode1);

	delete pNode1;
	pNode1 = nullptr;
}

// A list has multiple nodes, with a loop 
void Test3()
{
	ListNode* pNode1 = CreateListNode(1);
	ListNode* pNode2 = CreateListNode(2);
	ListNode* pNode3 = CreateListNode(3);
	ListNode* pNode4 = CreateListNode(4);
	ListNode* pNode5 = CreateListNode(5);

	ConnectListNodes(pNode1, pNode2);
	ConnectListNodes(pNode2, pNode3);
	ConnectListNodes(pNode3, pNode4);
	ConnectListNodes(pNode4, pNode5);
	ConnectListNodes(pNode5, pNode3);

	Test("Test3", pNode1, pNode3);

	delete pNode1;
	pNode1 = nullptr;
	delete pNode2;
	pNode2 = nullptr;
	delete pNode3;
	pNode3 = nullptr;
	delete pNode4;
	pNode4 = nullptr;
	delete pNode5;
	pNode5 = nullptr;
}

// A list has multiple nodes, with a loop 
void Test4()
{
	ListNode* pNode1 = CreateListNode(1);
	ListNode* pNode2 = CreateListNode(2);
	ListNode* pNode3 = CreateListNode(3);
	ListNode* pNode4 = CreateListNode(4);
	ListNode* pNode5 = CreateListNode(5);

	ConnectListNodes(pNode1, pNode2);
	ConnectListNodes(pNode2, pNode3);
	ConnectListNodes(pNode3, pNode4);
	ConnectListNodes(pNode4, pNode5);
	ConnectListNodes(pNode5, pNode1);

	Test("Test4", pNode1, pNode1);

	delete pNode1;
	pNode1 = nullptr;
	delete pNode2;
	pNode2 = nullptr;
	delete pNode3;
	pNode3 = nullptr;
	delete pNode4;
	pNode4 = nullptr;
	delete pNode5;
	pNode5 = nullptr;
}

// A list has multiple nodes, with a loop 
void Test5()
{
	ListNode* pNode1 = CreateListNode(1);
	ListNode* pNode2 = CreateListNode(2);
	ListNode* pNode3 = CreateListNode(3);
	ListNode* pNode4 = CreateListNode(4);
	ListNode* pNode5 = CreateListNode(5);

	ConnectListNodes(pNode1, pNode2);
	ConnectListNodes(pNode2, pNode3);
	ConnectListNodes(pNode3, pNode4);
	ConnectListNodes(pNode4, pNode5);
	ConnectListNodes(pNode5, pNode5);

	Test("Test5", pNode1, pNode5);

	delete pNode1;
	pNode1 = nullptr;
	delete pNode2;
	pNode2 = nullptr;
	delete pNode3;
	pNode3 = nullptr;
	delete pNode4;
	pNode4 = nullptr;
	delete pNode5;
	pNode5 = nullptr;
}

// A list has multiple nodes, without a loop 
void Test6()
{
	ListNode* pNode1 = CreateListNode(1);
	ListNode* pNode2 = CreateListNode(2);
	ListNode* pNode3 = CreateListNode(3);
	ListNode* pNode4 = CreateListNode(4);
	ListNode* pNode5 = CreateListNode(5);

	ConnectListNodes(pNode1, pNode2);
	ConnectListNodes(pNode2, pNode3);
	ConnectListNodes(pNode3, pNode4);
	ConnectListNodes(pNode4, pNode5);

	Test("Test6", pNode1, nullptr);

	DestroyList(pNode1);
}

// Empty list
void Test7()
{
	Test("Test7", nullptr, nullptr);
}

int main(int argc, char* argv[])
{
	Test1();
	Test2();
	Test3();
	Test4();
	Test5();
	Test6();
	Test7();

	return 0;
}