No. 29 - Loop in List

No. 29 - Loop in List

Question 1: How to check whether there is a loop in a linked list? For example, the list in Figure 1 has a loop.

A node in list is defined as the following structure:

struct ListNode

{

    int       m_nValue;

    ListNode* m_pNext;

};

Analysis: It is a popular interview question. Similar to the problem to get the K^th node from end is a list, it has a solution with two pointers.

Two pointers are initialized at the head of list. One pointer forwards once at each step, and the other forwards twice at each step. If the faster pointer meets the slower one again, there is a loop in the list. Otherwise there is no loop if the faster one reaches the end of list.

The sample code below is implemented according to this solution. The faster pointer is pFast, and the slower one is pSlow.

bool HasLoop(ListNode* pHead)

{

    if(pHead == NULL)

        return false;

    ListNode* pSlow = pHead->m_pNext;

    if(pSlow == NULL)

        return false;

    ListNode* pFast = pSlow->m_pNext;

    while(pFast != NULL && pSlow != NULL)

    {

        if(pFast == pSlow)

            return true;

        pSlow = pSlow->m_pNext;

        pFast = pFast->m_pNext;

        if(pFast != NULL)

            pFast = pFast->m_pNext;

    }

    return false;

}

Question 2: If there is a loop in a linked list, how to get the entry node of the loop? The entry node is the first node in the loop from head of list. For instance, the entry node of loop in the list of Figure 1 is the node with value 3.

Analysis: Inspired by the solution of the first problem, we can also solve this problem with two pointers. 

Two pointers are initialized at the head of a list. If there are n nodes in the loop, the first pointer forwards n steps firstly. And then they forward together, at same speed. When the second pointer reaches the entry node of loop, the first one travels around the loop and returns back to entry node.

Let us take the list in Figure 1 as an example. Two pointers, P1 and P2 are firstly initialized at the head node of the list (Figure 2-a). There are 4 nodes in the loop of list, so P1 moves 4 steps ahead, and reaches the node with value 5 (Figure 2-b). And then these two pointers move for 2 steps, and they meet at the node with value 3, which is the entry node of the loop.

The only problem is how to get the numbers in a loop. Let go back to the solution of the first question. We define two pointers, and the faster one meets the slower one if there is a loop. Actually, the meeting node should be inside the loop. Therefore, we can move forward from the meeting node and get the number of nodes in the loop when we arrive at the meeting node again.

The following function MeetingNode gets the meeting node of two pointers if there is a loop in a list, which is a minor modification of the previous HasLoop:

ListNode* MeetingNode(ListNode* pHead)

{

    if(pHead == NULL)

        return NULL;

    ListNode* pSlow = pHead->m_pNext;

    if(pSlow == NULL)

        return NULL;

    ListNode* pFast = pSlow->m_pNext;

    while(pFast != NULL && pSlow != NULL)

    {

        if(pFast == pSlow)

            return pFast;

        pSlow = pSlow->m_pNext;

        pFast = pFast->m_pNext;

        if(pFast != NULL)

            pFast = pFast->m_pNext;

    }

    return NULL;

}

We can get the number of nodes in a loop of a list, and the entry node of loop after we know the meeting node, as shown below:

ListNode* EntryNodeOfLoop(ListNode* pHead)

{

    ListNode* meetingNode = MeetingNode(pHead);

    if(meetingNode == NULL)

        return NULL;

    // get the number of nodes in loop

    int nodesInLoop = 1;

    ListNode* pNode1 = meetingNode;

    while(pNode1->m_pNext != meetingNode)

    {

        pNode1 = pNode1->m_pNext;

        ++nodesInLoop;

    }

    // move pNode1

    pNode1 = pHead;

    for(int i = 0; i < nodesInLoop; ++i)

        pNode1 = pNode1->m_pNext;

    // move pNode1 and pNode2

    ListNode* pNode2 = pHead;

    while(pNode1 != pNode2)

    {

        pNode1 = pNode1->m_pNext;

        pNode2 = pNode2->m_pNext;

    }

    return pNode1;

}

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